Exercise 26.2
निम्नलिखित का प्रथम सिद्धान्त से अवकल गुर्णांक (अवकलन) निकाले
[Find the differential coefficient (derivative) of the following from first principles]
Question 1
(i) 10x
Sol :
Let f(x)=10x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{10(2+h)-10 x}{h}$
$=\lim _{h \rightarrow 0} \frac{10 x+10 h-10 x}{h}$
$=\lim _{h \rightarrow 0} \frac{10 h}{h}$
=10
$d\left(\frac{10 x}{dx}\right)=10$
(ii) x3
Sol :
Let f (x)=x3
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{h}$
$=\lim_{h\rightarrow 0}\frac{(x+h)^{3}-x^{3}}{(x+h)-0}$
=3.x3-1
=3x2
$\frac{d\left(x^{3}\right)}{d}=3 \cdot x^{2}$
(iii) 5x4
Sol :Let f(x)=5x4
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+4)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{5(x+4)^{4}-5 x^{4}}{4}$
$=5 \lim _{h \rightarrow 0} \frac{(x+h)^{4}-x^{4}}{(x+4)-x}$
=5.4x4-1
=20x3
=20x3
Question 2
(i) $\frac{1}{x}$
Sol :
Let f(x)$=\frac{1}{x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{x+h}-\frac{1}{x}}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{x-x-h}{(x+h) x}}{h}$
$=\lim_{h \rightarrow 0} \frac{-h}{h(x+h)x}$
$=\lim_{h \rightarrow 0} \frac{1}{(x+h) \cdot x}$
$=-\frac{1}{(x+0) \cdot x}$
$=-\frac{1}{x^{2}}$
(ii) $\frac{1}{x^{2}}$
Sol :
Let $f(x)=\frac{1}{x^{2}}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{x^{2}-(x+h)^{2}}{(x+h)^2.x^2}}{h}$
$=\lim _{h \rightarrow 0} \frac{x^{2}-x^{2}-2 x h-h^{2}}{h(x+h)^{2} \cdot
x^{2}}$
$=\lim_{h\rightarrow 0} \frac{-h(2 x+h)}{h(x+4)^{2} \cdot x^{2}}$
$=\frac{-(2 x+0)}{(x+0)^{2} \cdot x^{2}}$
$=\frac{-2 x}{x^{4}}=\frac{-2}{x^{3}}$
(iii) (-x)-1
Sol :
Let $f(x)=(-x)^{-1}=\frac{1}{-x}$ या $\frac{-1}{x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{1}{x+h}+\frac{1}{x}}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{-x+x+h}{(x+h) \cdot x}}{h}$
$=\lim_{h \rightarrow 0} \frac{h}{h(x+4) \cdot x}$
$=\frac{1}{(x+0) \cdot x}=\frac{1}{x^{2}}$
Question 3
$\frac{1}{\sqrt{2 x}}$
Sol :
माना $f(x)=\frac{1}{\sqrt{2 x}}$
$f'(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{\sqrt{2(x+h)}}-\frac{1}{\sqrt{2
x}}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\sqrt{2}
x-\sqrt{2(x+4)}}{\sqrt{2(x+4) \cdot \sqrt{2 x}}}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2 x}-\sqrt{2(x+h)}}{h\sqrt{2(x+4)
\sqrt{2 x}}}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2 x}-\sqrt{2(x+6)}}{h\sqrt{2(x+2)
\sqrt{2 x}}} \times \frac{\sqrt{2 x}+\sqrt{2(x+6)}}{\sqrt{2
x+\sqrt{2(x+1)}}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{2} x)^{2}-(\sqrt{2(x+h)})^2}{h\times
2 \sqrt{(x+h)x} \left[ \sqrt{2 x}+\sqrt{2(x+h)}\right]}$
$=\lim _{h \rightarrow 0} \frac{2 x-2 x-2 h}{2 h \sqrt{(x+4) x}[\sqrt{2
x}+\sqrt{2(x+4)}]}$
$=\lim _{h \rightarrow 0} \frac{-2 h}{2 h \sqrt{x+h}] \cdot x[\sqrt{2
x}+\sqrt{2(x+h)}]}$
$=\frac{-1}{\sqrt{(x+0) \cdot x}[\sqrt{2 x}+\sqrt{2(x+0)}]}$
$=\frac{-1}{x \cdot 2 \sqrt{2 x}}$
$=\frac{-1}{2 \sqrt{2} \cdot x \cdot x^{1/2}}$
$=\frac{-1}{2 \sqrt{2} x^{3/ 2}}$
Question 4
(i) $\sqrt{2 x-a}$
Sol :
माना $f(x)=\sqrt{2 x-a}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\sqrt{2(x+h)-a}-\sqrt{2 x-a}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{2(x+h)-a}-\sqrt{2 x-a} }{h}\times
\dfrac{\sqrt{2(x+1)-a}+\sqrt{2 x-a}}{\sqrt{2(x+h)-a}+\sqrt{2 x-a}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{2(x+h)-a})^{2}-(\sqrt{2
x-a})^{2}}{h[\sqrt{2(x+h)-a}+\sqrt{2 x-a}]}$
$=\lim _{h \rightarrow 0} \frac{2 x+2 h-a-2 x+a}{h[\sqrt{2(x+h)-a}+\sqrt{2
x-a}]}$
$=\lim _{h \rightarrow 0} \frac{2h}{h[\sqrt{2(x+h)-a}+\sqrt{2 x-a}]}$
$=\frac{2}{\sqrt{2(x+0)-a}+\sqrt{2 x-a}}$
$=\frac{2}{2 \sqrt{2 x-a}}$
$=\frac{1}{\sqrt{2 x-a}}$
(ii) $\sqrt{x}$
Sol :
माना $f(x)=\sqrt{x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$\lim_{h\rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \times
\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$
$=\lim_{h \rightarrow 0}
\frac{(\sqrt{x+4})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x}]}$
$=\lim _{h \rightarrow 0} \frac{x+h-x}{h[\sqrt{x+h}+\sqrt{x}]}$
$=\lim _{h \rightarrow 0} \frac{h}{h[\sqrt{x+h}+\sqrt{x}]}$
$=\frac{1}{\sqrt{x+0}+\sqrt{x}}$
$=\frac{1}{2 \sqrt{x}}$
Question 5
(i) x8+x3
Sol :
माना f(x)=x8+x3
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{(x+4)^{8}+(x+h)^{3}-x^{3}-x^{3}}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+1)^{8}-x^{3}+(x+h)^{3}-x^{3}}{(x+h)-x}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{8}-x^{8}}{(x+1)-x}+\lim _{h
\rightarrow 0} \frac{(x+h)^{3}-x^{3}}{(x+h)-x}$
=8x7+3x2
(ii) x3+3x2+5
Sol :
माना f(x)=x3+3x2+5
$f^{\prime}(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{3}+3(x+h)^{2}+5-x^3-3x^2-5}{h}$
$=\lim _{h \rightarrow 0}
\frac{(x+4)^{3}-x^{3}+3\left[(x+h)^{2}-x^{2}\right]}{(x+h)-x}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{3}-x^{3}}{(x+h)-x}+3 \lim _{h
\rightarrow 0} \frac{(x+h)^{2}-x^{2}}{(x+h)-x}$
=3x2+3×2x1
=3x2+6x
(iii) (x-1)(x-2)
(v) $x+\frac{1}{x}$
Sol :
माना $f(x)=x+\frac{1}{x}$
$f^{ \prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+5)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{x+h+\frac{1}{x+h}-x-\frac{1}{x}}{h}$
$=\lim _{h \rightarrow 0} \frac{h}{h}+\lim _{h \rightarrow 0}
\frac{\frac{1}{x+h}-\frac{1}{x}}{h}$
$=1+\lim _{h \rightarrow 0} \dfrac{\frac{x-x-h}{(x+h) x}}{h}$
$=1+\lim _{h \rightarrow 0} \frac{-h}{h(x+h) \cdot x}$
$=1+\frac{-1}{(x+0) x}$
$=1-\frac{1}{x^{2}}$
Question 6
(i) $\frac{x+1}{x-1}$
Sol :
माना $f(x)=\frac{x+1}{x-1}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1}}{h}$
$=\lim _{h \rightarrow 0}
\dfrac{\frac{(x-1)(x+h+1)-(x+1)(x+h-1)}{(x+h-1)(x-1)}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{x^{2}+hx+x-x-h-1-x^{2}-h
x+x-x-h+1}{h(x+h-1)(x-1)}$
$=\lim _{h \rightarrow 0} \frac{-2h}{h(x+h-1)(x-1)}$
$=\frac{-2}{(x+0-1)(x-1)}$
$=\frac{-2}{(x-1)^{2}}$
Question 7
sin 4x
Sol :
माना f(x)=sin 4x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin 4(x+h)-\operatorname{sin} 4 x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{2 \sin \frac{4 x+4 h-4 x}{2} \cos \frac{4
x+4h+4x}{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{2 \sin 2 h \times 2 \cdot \cos \frac{8
x+4h}{2}}{2h}$
$=4 \lim _{h \rightarrow 0} \frac{\sin 2 h}{2 h} \cdot \lim _{h \rightarrow
0} \cos (4 x+24)$
=4×1×cos(4x+2(0))
=4cos4x
Question 8
cos2x
Sol :
माना f(x)=cos2x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos 2(x+h)-\cos 2 x}{h}$
$=\lim_{h\rightarrow 0} \dfrac{-2 \sin \frac{2 x+2 h+2x}{2} \sin \frac{2
x+2h-2 x}{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{-2 \cdot \sin \frac{4 x+2h}{2} \sin h}{h}$
$=-2 \lim _{h \rightarrow 0} \sin (2 x+h) \cdot \lim _{h \rightarrow 0}
\frac{\sin {h}}{h}$
=-2sin(2x+0)×1
=-2sin2x
Question 10
tan kx
Sol :
माना f(x)=tan kx
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\tan k(x+h)-\tan k x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\sin k(x+h)}{\cos k(x+h)}-\frac{\sin
kx}{\cos kx}}{h}$
$=\lim_{h\rightarrow 0}\dfrac{\frac{\sin k(x+h).\cos kx-\cos k(x+h)\sin
kx}{\cos k(x+h) \cdot \cos kx}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (k x+k h-k x)}{h \cos k(x+h) \cdot \cos
k x}$
$=\lim _{h \rightarrow 0} \frac{\sin k h}{h \cdot \cos k(x+h) \cdot \cos
kx}$
$=\lim _{h \rightarrow 0} \frac{\sin k h}{k h} \times k ~\lim_{h\rightarrow
0}\frac{1}{\cos k(x+h).\cos kx}$
$=1 \times k \times \frac{1}{\cos k(x+0) \cdot \cos kx}$
$=\frac{k}{\cos ^{2} k x}$
=ksec2kx
Question 11
sin ax
Sol :
माना f(x)=sin ax
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{\sin a(x+h)-\sin a x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{2 \sin \frac{ax+ah-ax}{2}\cdot \cos
\frac{ax+ah+ax}{2}}{h}$
$=\lim_{h \rightarrow 0} \frac{\sin \frac{a h}{2}}{\frac{ah}{2}}\times a
.\lim_{h\rightarrow 0}\frac{\cos \frac{2ac+ah}{2}}{2}$
$=1 \times a \cdot \cos \frac{2 a x+a(0)}{2}$
$=a \cdot \cos \frac{2a x}{2}$
=acos ax
Question 12
sec(2x+3)
Sol :
माना f(x)=sec(2x+3)
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sec [2(x+h)+3]-\sec (2 x+3)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{1}{\cos [2(x+h)+3]}-\frac{1}{\cos (2
x+3)}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\cos (2 x+3)-\cos [2(x+h)+3]}{\cos
[2(x+h)+3] \cdot \cos (2 x+3)}}{h}$
$=\lim_{h\rightarrow 0} \dfrac{2 \sin \frac{2 x+3+2 x+2h+3}{2}+3 \sin
\frac{2 x+2h+3-2 x}{2}}{h \cos \left[2(x+h)+3\right]\cos(2x+3)}$
$=\lim_{h\rightarrow 0} \dfrac{2 \sin \frac{2 x+3+2 x+2h+3}{2} \sin \frac{2
x+2h+3-2 x-3}{2}}{h \cos \left[2(x+h)+3\right]\cos(2x+3)}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{4 x+2 h+6}{2} \sin
\frac{2h}{2}}{h \cdot \cos (2(x+4)+3] \cdot \cos (2 x+3)}$
$=2 \lim _{h \rightarrow 0} \frac{{\sin (2 x+h+3)}}{\cos [2(x+4)+3] \cdot
\cos (2 x+3)}\lim{\frac{\sin h}{h}}$
$=2 \cdot \frac{\sin (2 x+0+3)}{\cos [2(x+0)+3] \cos (2 x+3)} \times 1$
$=\frac{2\sin (2 x+3)}{\cos (2 x+3) \cdot \cos (2 x+3)}$
=2sec(2x+3)tan(2x+3)
Question 13
$\tan \frac{x^{\circ}}{2}$
Sol :
$=\tan \frac{x}{2} \times \frac{\pi}{180}$
$=\tan \frac{\pi x}{360}$
माना $f(x)=\tan \frac{\pi x}{360}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\tan \frac{\pi(x+h)}{360}-\tan \frac{\pi
x}{360}}{h}$
$\lim_{h\rightarrow 0}\dfrac{\frac{\sin \frac{\pi(x+h)}{360}}{\cos
\frac{\pi(x+h)}{360}}-\frac{\sin \frac{\pi x}{360}}{\cos \frac{\pi
x}{360}}}{h}$
$=\lim _{h \rightarrow 0}\dfrac{\frac{\sin \frac{\pi(x+h)}{360} \cdot \cos
\frac{\pi x}{360}-\cos \pi \left(\frac{x+h}{360}\right).\sin \frac{\pi
x}{360}}{\cos \frac{\pi(x+1)}{360} \cos \frac{\pi 2}{360}}}{h}$
$\lim_{h\rightarrow 0} \dfrac{\sin \left(\frac{\pi x}{360}+\frac{\pi
h}{360}-\frac{\pi x}{360}\right)}{h.\cos \frac{\pi(x+4)}{360} \cdot \cos
\frac{\pi x}{360}}$
$=\lim _{h \rightarrow 0} \frac{\sin \frac{\pi h}{360}}{\frac{\pi}{360} h
}\times \frac{\pi}{360}~ \lim _{h \rightarrow
0}\dfrac{1}{\cos\frac{\pi(x+h)}{360}.\cos \frac{\pi x}{360}}$
$=1 \times \frac{\pi}{360} \times \frac{1}{\cos \pi \frac{(x+0)}{360} \cdot
\cos \frac{\pi x}{360}}$
$=\frac{\pi}{360} \times \frac{1}{\cos^{2} \frac{\pi x}{360}}$
$=\frac{\pi}{360} \sec ^{2} \frac{\pi x}{360}$
$=\frac{\pi}{360} \cdot \sec ^{2}\left(\frac{\pi x}{360} \times
\frac{180}{\pi}\right)^{\circ}$
$=\frac{\pi}{360} \sec ^{2} \frac{x^{0}}{2}$
Question 15
cos2x
Sol :
माना f(x)=cos2x
$f^{\prime}(x)=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos ^{2}(x+h)-\cos ^{2} x}{h}$
$=\lim _{h \rightarrow 0} \frac{\operatorname{sin}(x+x+h) \cdot \sin (x-x-h)}{h}$
$=\lim_{h \rightarrow 0} \dfrac{\sin (2 x+h)-(-\sin h)}{h}$
$=-\lim _{h \rightarrow 0} \sin (2 x+h).\lim _{h \rightarrow 0} \frac{\sin h}{h}$
=-sin(2x+0)×1
=-sin(2x+0)
Question 15
cos x2
Sol :
माना f(x)=cos x2
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos (x+h)^{2}-\cos x^{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos \left(x^{2}+2 h x+h^{2}\right)-\cos x^{2}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{-2 \sin \frac{x^{2}+2h x+h^{2}+x^{2}}{2}\sin \frac{x^2+2hx+h^2-x^2}{2}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{-2 \sin \left(\frac{2 x^{2}+2 h x+h^{2}}{2}\right) \sin \frac{(2 x+h) h}{2}}{h}$
$=-\lim _{h \rightarrow 0} \sin \left(\frac{2 x^{2}+2 h x+h^{2}}{2}\right) \cdot \lim _{h \rightarrow 0} \dfrac{\frac{\sin (2 x+h)h}{2}}{\frac{h}{2}}$
$=-\lim _{h \rightarrow 0} \frac{\sin \left(2 x^{2}+2 h x+h^{2}\right)}{2} \cdot \lim _{h \rightarrow 0}\dfrac{\sin \frac{(2x+h)h}{2}}{(2x+h)\frac{h}{2}}\times(2x+h)$
$=-\frac{\sin \left(2 x^{2}+2 \times 0 \cdot x+0^{2}\right)}{2}\times 1\times (2x+0)$
$=-\sin \frac{2 x^{2}}{x} \cdot 2 x$
=-2x sin x2
Question 17
sin (x2+2)
माना f(x)=sin (x2+2)
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \sin \frac{\left[(x+h)^{2}+2\right]-\sin \left(x^{2}+2\right)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\sin \left(x^{2}+2 h x+h^{2}+2\right)-\sin \left(x^{2}+2\right)}{h}$
$\lim_{h\rightarrow 0}\dfrac{2 \sin \frac{x^{2}+2 h x+h^{2}+2-x^{2}-2}{2} \cdot \cos \frac{x^{2}+2 h x+h^{2}+2+x^2+2}{2}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{2 \sin (2 x+h) h}{2} \cos \frac{2 x^{2}+2 h x+h^{2}+4}{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+h}{2}\right){h}}{(2 x+h)\frac{h}{2}} \times(2 x+4).\lim_{h\rightarrow 0}\cos\dfrac{(2x^2+2hx+h^2+4)}{2}$
$=1 \times(2 x+0) \cdot \cos \left(\frac{2 x^{2}+2 \times 0 x+0^{2}+4}{2}\right)$
$=2 x \cos \left(\frac{2 x^{2}+4}{2}\right)$
=2xcos(x2+2)
Question 18
sin3x
Sol :
माना f(x)=sin3x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \dfrac{ f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin^{3}(x+h)-\sin^{3} x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{[\sin (x+h)-\sin x][\sin ^2 (x+h)+\sin (x+h)\sin x+\sin^2 x]}{h}$
$=\lim _{h \rightarrow 0}\dfrac{2 \sin \frac{x+h-x}{2}}{h} \lim _{h \rightarrow 0}\left[\sin ^2 (x+h)+\sin (x+h).\sin x+\sin ^2 x\right]$
$=\lim_{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \lim _{h \rightarrow 0} \cos \left(\frac{2 x+h}{2}\right)\lim_{h\rightarrow 0}\left[\sin ^2 (x+h)+\sin(x+h).\sin x+\sin ^2 x\right]$
$=1 \times \cos \left(\frac{2 x+0}{2}\right) \left[\sin ^2 (x+0)+\sin(x+0).\sin x+\sin ^2 x\right]$
=cos x.3sin3x
=3sin2x.cos x
Question 19
$\sqrt{\tan x}$
Sol :
माना f(x)=$\sqrt{\tan x}$
$f^{\prime}(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f{(x)}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\tan (x+h)}- \sqrt{\tan x}}{h}$
$=4_{\rightarrow 0} \dfrac{\sqrt{\tan (x+h)}-\sqrt{\tan x}}{h}\times \dfrac{\sqrt{\tan (x+h)+\sqrt{\tan x}}}{\sqrt{\tan (x+h)+\sqrt{\tan x}}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{\tan (x+h)})^{2}-(\sqrt{\tan x})^{2}}{h[\sqrt{\tan (x+h)}+\sqrt{\tan x}]}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\sin (x+4)}{\cos (x+4)}-\frac{\sin x}{\cos x}}{h[\sqrt{\tan (x+h)}+\sqrt{\tan x}]}$
$\lim_{h\rightarrow 0}\dfrac{\frac{\sin (x+h) \cdot \cos x-\cos (x+h)\sin x}{\cos (x+h) \cos x}}{h[\sqrt{\tan (x+h)}+\sqrt{\tan x}]}$
$=\lim _{h \rightarrow 0} \frac{\sin (x+h-x)}{\cos (x+4) \cdot \cos x \cdot h\left[\sqrt{\tan (x+h)}+\sqrt{tan x}\right]}$
$=\lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \dfrac{1}{\cos (x+h).\cos x[\sqrt{\tan (x+h)+\sqrt{\tan x}}]}$
$=1 \times \frac{1}{\cos (x+0) \cos x[\sqrt{\tan (x+0)}+\sqrt{\tan x}}$
$=\frac{1}{\cos ^{2} x \cdot 2 \sqrt{\tan x}}$
$=\frac{\sec ^{2} x}{2 \sqrt{\tan x}}$
Question 20
x2cos x
Sol :
माना f(x)=x2cos x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h)^{2} \cdot \cos (x+h)-x^{2} \cos x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\left(x^{2}+2 x h+h^{2}\right) \cdot \cos (x+h)-x^{2} \cos x}{h}$
$=\lim _{h \rightarrow 0} \dfrac{x^{2} \cos (x+h)-x^{2} \cos x+2 x h \cos (x+4)+h^{2} \cos (x+h)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{x^{2}[\cos (x+h)-\cos x]}{h}+\lim _{h \rightarrow 0} \frac{2 x h (x+h)}{h}+\lim_{h\rightarrow 0} \dfrac{h^2\cos(x+h)}{h}$
$=\lim _{h \rightarrow 0} \frac{ x^{2}\left[-2 \sin \frac{x+h+x}{2} \sin \frac{x+4-x}{2}\right]}{h}+2x\cos(x+0)+0.\cos(x+0)$
$=\lim _{h \rightarrow 0} x^{2} \dfrac{\left[-2 \sin \left(\frac{2 x+6}{2}\right) \sin \frac{h}{2}\right]}{h}+2x\cos x$
$=-\lim _{h \rightarrow 0} x^{2} \sin \left(\frac{2 x+h}{2}\right).\lim_{h\rightarrow 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}+2x\cos x$
$=-x^{2} \lim _{h \rightarrow 0} \sin \left(x+\frac{h}{2}\right) \cdot 1+2 x \cos x$
$=-x^{2}\sin \left(x+\frac{0}{2}\right)+2 x \cos x$
=-x2sinx+2xcosx
=2xcosx-x2sinx
Question 21
(i) cos(3x+2)
Sol :
माना f(x)=cos(3x+2)
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos [3(x+h)+2]-\cos (3 x+2)}{h}$
$=\lim_{h\rightarrow 0}\dfrac{-2 \sin \frac{3 x+3h+2+3 x+2}{2}\sin\frac{3 x+3h+2-3x-2}{2}}{h}$
$=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{6 x+3 h+4}{2} \cdot \sin \frac{3h}{2}}{h}$
$=-\lim _{h \rightarrow 0} \sin \left(\frac{6 x+3 h+4}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{3 h}{2}}{ \frac{3h}{2}} \times {3}$
$=-\sin \left(\frac{6 x+3(0)+4}{2}\right) \times 1 \times 3$
$=-\sin \frac{2(3 x+2)}{2} \times 3$
=-3sin(3x+2)
(ii) $\cos \left(x-\frac{\pi}{8}\right)$
Sol :
माना f(x)=$\cos \left(x-\frac{\pi}{8}\right)$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\cos \left(x+h-\frac{\pi}{8}\right)-\cos \left(x-\frac{\pi}{8}\right)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{-2 \sin \frac{x+h-\frac{\pi}{8}+x-\frac{\pi}{8}}{2}.\sin \dfrac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2}}{h}$
$=-\lim _{h \rightarrow 0} \sin \left(\frac{2 x+h-\frac{\pi}{4}}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=-\sin \left(\frac{2 x+0-\frac{\pi}{4}}{2}\right) \times 1$
$=-\sin \left(x-\frac{\pi}{8}\right)$
Question 22
(i) sin(x+1)
Sol :
माना f(x)=sin(x+1)
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (x+h+1)-\sin (x+1)}{4}$
$=\lim _{h \rightarrow 0}\dfrac{ 2 \sin \frac{x+h+1-x-1}{2} \cdot \cos\frac{ \frac{x+h+1+x+1}{2}}{4}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \lim _{h \rightarrow 0} \cos \left(\frac{2 x+h+2}{2}\right)$
$=1 \times \cos \left(\frac{2 x+0+2}{2}\right)$
=cos(x+1)
(ii) sin(2x+1)
Sol :
Question 23
cos3x
Sol :
माना f(x)=cos3x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos ^{3}(x+h)-\cos ^{3} x}{h}$
$=\lim _{h \rightarrow 0}\dfrac{[\cos (x+h)-\cos x]\left[\cos ^{2}(x+h)+\cos (x+h) \cdot \cos x+\cos ^{2}x\right]}{h}$
$=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{x+h+x}{2} \sin \frac{x+h-x}{2}}{h}.\lim_{h\rightarrow 0}\left[\cos^2 (x+h)+\cos(x+h).\cos x+\cos^2 x\right]$
$=-\lim _{h \rightarrow 0} \sin \left(\frac{2 x+h}{2}\right) \cdot \lim _{h \rightarrow 0} \frac{ \sin\frac{h}{2}}{\frac{h}{2}}.\lim_{h\rightarrow 0}\left[\cos^2 (x+h)+\cos(x+h).\cos x+\cos^2 x\right]$
$=-\sin \left(\frac{2 x+0}{2}\right) \times 1\times\left[\cos^2 (x+0)+\cos(x+0).\cos x+\cos^2 x\right]$
$=-\sin \frac{2x}{2} \times 3 \cos^2 x$
=-3cos3x.sinx
Question 24
tan(ax+b)
Sol :
माना f(x)=tan(ax+b)
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\tan [a(x+h)+b]-\tan (a x+b)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\sin [a(x+h)+b]}{\cos [a(x+h)+b]}-\frac{\sin (a x+b)}{ \cos (a x+b)}}{h}$
$=\lim_{h\rightarrow 0}\dfrac{\frac{\sin [a(x+h)+b] \cdot \cos (a x+b)-\cos [a(x+h)+b].\sin(ax+b)}{\cos [a(x+h)+b] \cdot \cos (a x+b)}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (a x+a h+b-ax-b)}{h \cos [a(x+h)+b] \cdot \cos (a x+b)}$
$=\lim_{h \rightarrow 0} \frac{\sin a h}{a h}\times a.\lim_{h\rightarrow 0}\frac{1}{\cos[a(x+h)+b].cos(ax+b)}$
$=1 \times a \times \frac{1}{\cos [a(x+0)+b] \cdot \cos \left(ax+b\right)}$
$=\frac{a}{\cos ^{2}(ax+b)}$
=asec2(ax+b)
Question 25
xsinx
Sol :
माना f(x)=xsinx
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{(x+h) \sin (x+h)-x \sin x}{h}$
$=\lim _{h \rightarrow 0} \frac{x \sin (x+h)-x \sin x+h \sin(x+h)}{h}$
$=\lim _{h \rightarrow 0} \frac{x[\sin (x+h)-\sin x]}{h}+\lim_{h\rightarrow 0}\frac{h\sin (x+h)}{h}$
$=x \lim _{h \rightarrow 0} \frac{2 \sin \frac{x +-x}{2} \cos \frac{x+h+x}{2}}{h}+\sin (x+0)$
$=x \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \lim _{h \rightarrow 0} \cos \left(\frac{2 x+h}{2}\right)+\sin x$
$=x \cdot 1 \cdot \cos \left(\frac{2 x+0}{2}\right)+\sin x$
=xcos x+sin x
Question 26
$\sqrt{\cos x}$
Sol :
माना f(x)=$\sqrt{\cos x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\cos (x+h)}-\sqrt{\cos x}}{h}\times \frac{\sqrt{\cos (x+h)}+\sqrt{\cos x}}{\sqrt{\cos (x+h)}+\sqrt{\cos x}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{\cos (x+h)})^{x}-(\sqrt{\cos x})^{2}}{h[\sqrt{\cos (x+h)+\sqrt{\cos x}]}}$
$=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h(\sqrt{\cos (x+h)}+\sqrt{\cos x}]}$
$=\lim_{h\rightarrow 0}\dfrac{-2 \sin\frac{x+h+x}{2} \cdot \sin \frac{x+h-x}{2}}{h[\sqrt{\cos (x+h)}+\sqrt{\cos x}]}$
$=-\lim _{h \rightarrow 0} \frac{\sin \left(\frac{2 x+1}{2}\right)}{\sqrt{\cos (x+h)}+\sqrt{\cos x}}.\lim_{h\rightarrow 0}\dfrac{\sin\frac{h}{2}}{\frac{h}{2}}$
$=\frac{-\sin \left(\frac{2 x+0}{2}\right)}{\sqrt{\cos (x+0)+\sqrt{\cos x}}}\times 1$
$=\frac{-\sin x}{\sqrt{\cos x}+\sqrt{\cos x}}$
$=\frac{-\operatorname{sin} x}{2 \sqrt{\cos x}}$
Question 27
$\sqrt{\sec x}$
Sol :
माना f(x)=$\sqrt{\sec x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\sec (x+h)}-\sqrt{\sec x}}{h}\times \frac{\sqrt{\sec (x+h)}+\sqrt{\sec x}}{\sqrt{\sec (x+h)+\sqrt{\operatorname{sec} x}}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{\sec (x+h)})^{2}-(\sqrt{\sec x})^{2}}{h\sqrt{\sec (x+h)}+\sqrt{\sec x})}$
$=\lim _{h \rightarrow 0} \frac{\frac{1}{\cos (x+h)}-\frac{1}{\cos x}}{h[\sqrt{\sec (x+h)}+\sqrt{\sec x}]}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\cos x- \cos (x+h)}{\cos(x+h).\cos x}}{h(\sqrt{\operatorname{sec}(x+h)}+\sqrt{\sec x}]}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{x+x+4}{2}.\sin\frac{x+h-x}{2}}{h[\sqrt{\sec x+h}]+\sqrt{\sec x] \cdot \cos (x+h) \cdot \cos x}}$
$=\lim_{h \rightarrow 0} \frac{\sin \frac{2 x+h}{2}}{[\sqrt{\sec (x+h)}+\sqrt{\operatorname{sec} x}] \cdot \cos (x+h) \cdot \cos x}\lim_{h\rightarrow 0}\dfrac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=\lim_{h \rightarrow 0} \frac{\sin \frac{2 x+h}{2}}{[\sqrt{\sec (x+h)}+\sqrt{\operatorname{sec} x}] \cdot \cos (x+h) \cdot \cos x}\times 1$
$=\frac{\sin x}{2 \sqrt{\sec x} \cdot \cos x \cdot \cos x}$
$=\frac{\sec x+\tan x}{2 \sqrt{\sec x}}$
Question 28
$\sqrt{\sin 3 x}$
Sol :
माना f(x)=$\sqrt{\sin 3 x}$
माना f(x)=$\sqrt{\sin 3 x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\sin 3(x+h)}-\sqrt{\sin 3 x}}{h}$
$=\lim _{h \rightarrow 0} \frac{\sqrt{\operatorname{sin} 3(x+h)}-\sqrt{\sin 3 x}}{h}\times \frac{\sqrt{\sin 3(x+h)}+\sqrt{\sin 3 x}}{\sqrt{\sin 3(x+h)}+\sqrt{\sin 3 x}}$
$=\lim _{h \rightarrow 0} \frac{(\sqrt{\sin 3(x+h)})^{2}-(\sqrt{\sin 3x})^2}{h[\sqrt{\sin 3(x+h)}+\sqrt{\sin 3 x}]}$
$=\lim _{h \rightarrow 0} \frac{\sin (3 x+3 h)-\sin 3 x}{h[\sqrt{\sin 3(x+h)}+\sqrt{\sin 3 x}]}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{3 x+3 h-3 x}{2} \cdot \cos \frac{3 x+3 h+3x}{2}}{h \sqrt{\sqrt{\sin 3(x+h)}+\sqrt{\sin 3 x}]}}$
$=\lim _{h \rightarrow 0} \frac{\sin \frac{3 h}{2}}{\frac{3}{2}h} \times 3 \lim_{h\rightarrow 0} \dfrac{\cos \frac{6x+3h}{2}}{\sqrt{\sin 3(x+h)}+\sqrt{\sin 3x}}$
$=1 \times 3 \cdot \frac{\cos \frac{6 x+3(0)}{2}}{\sqrt{\sin 3(x+0)}+\sqrt{\sin 3 x}}$
$=\frac{3 \cdot \cos 3 x}{2 \sqrt{\sin 3 x}}$
Question 29
$\sin \left(x^{2}+1\right)$
Sol :
माना f(x)=$\sin \left(x^{2}+1\right)$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\tan \left[(x+h)^{2}+1\right]-\tan \left(x^{2}+1\right)}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{\sin \left[(x+h)^{2}+1\right]}{\cos [(x+h)^2+1]}-\frac{\sin \left(x^{2}+1\right)}{\cos \left(x^{2}+1\right)}}{h}$
$=\lim_{h\rightarrow 0}\dfrac{\frac{\sin \left[(x+h)^{2}+1\right] \cdot \cos\left(x^{2}+1\right)-\cos \left[(x+h)^{2}+1\right] \cdot \sin \left(x^{2}+1\right)}{\cos \left[(x+h)^{2}+1\right] \cdot \cos \left(x^{2}+1\right)}}{h}$
$=\lim _{h \rightarrow 0} \frac{\left.\sin \left[x^{x}+2 h x+h^{2}+1-x^{2}-1\right.\right]}{h \cdot \cos \left[(x+h)^{2}+1\right] \cdot \cos \left(x^{2}+1\right)}$
$=\lim _{h \rightarrow 0} \frac{\sin (2 x+h) \cdot h}{(2 x+4) \cdot h}(2 x+4)\times \lim_{h\rightarrow 0}\frac{1}{\cos [(x+h)^2+1].\cos (x^2+1)}$
$=1 \times(2 x+0)\times \frac{1}{\left[\cos (x+0)^{2}+1\right] \cdot \cos \left(x^{2}+1\right)}$
$=\frac{2 x}{\cos ^{2}\left(x^{2}+1\right)}$
$=2 x \cdot \sec ^{2}\left(x^{2}+1\right)$
Question 30
$\cos \sqrt{x}$
Sol :
माना f(x)=$\cos \sqrt{x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\cos \sqrt{x+h}-\cos \sqrt{x}}{(x+h)-x}$
$=\lim _{h \rightarrow 0} \frac{-2 \sin \frac{\sqrt{x+h}+\sqrt{x}}{2} \cdot \sin \frac{\sqrt{x+h}-\sqrt{x}}{2}}{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}$
$=-\lim _{h \rightarrow 0} \frac{2\sin \frac{\sqrt{x+h}+\sqrt{x}}{2}\times \sin \frac{\sqrt{x+h}-\sqrt{x}}{2}}{(\sqrt{x+h}+\sqrt{x})(\sqrt{x+h}-\sqrt{x})}$
$=-\lim _{h \rightarrow 0} \frac{\sin \sqrt{x+h}+\sqrt{x}}{\frac{2}{(\sqrt{x+h}+\sqrt{x})}}\lim_{h\rightarrow 0} \dfrac{\sin \frac{\sqrt{x+h}-\sqrt{x}}{2}}{\frac{\sqrt{x+h}-\sqrt{x}}{2}}$
$=\frac{\sin \frac{\sqrt{x+0}+\sqrt{x}}{2}}{\sqrt{x+0}+\sqrt{x}} \times 1$
$=-\frac{\sin \frac{2\sqrt{x}}{2}}{2 \sqrt{x}}$
$=-\frac{\sin \sqrt{x}}{2 \sqrt{x}}$
Question 32
$\frac{\cos x}{x}$
Sol :
माना f(x)=$\frac{\cos x}{x}$
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\frac{\cos (x+h)}{x+h}-\frac{\cos x}{x}}{h}$
$=\lim _{h \rightarrow 0} \dfrac{\frac{x \cos (x+h)-x \cos x-h \cos x}{(x+h)x}}{h}$
$=\lim _{h \rightarrow 0} \frac{x[\cos (x+h)-\cos x]-h \cos x}{h(x+h) \cdot x}$
$=\lim _{h \rightarrow 0} \dfrac{x\left[-2 \sin \frac{x+h+x}{2} \sin \frac{x+h-x}{2}\right]-h \cos x}{h(x+h)x}$
$=\lim _{h \rightarrow 0} \frac{-2 x \sin \frac{2 x+3}{2} \sin \frac{h}{2}-h \cos x}{h(x+h) \cdot x}$
$=\lim _{h \rightarrow 0} \frac{-2 x \sin \frac{2 x+h}{2} \sin \frac{h}{2}}{h(x+h) x}-\lim _{h \rightarrow 0} \frac{h \cos x}{h \cdot(x+h)x}$
$=-x \lim _{h \rightarrow 0} \frac{\sin \frac{2 x+h}{2}}{(x+h) \cdot x}\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}-\lim _{h \rightarrow 0} \frac{\cos x}{(x+h) \cdot x}$
$=-x \dfrac{\sin \frac{2 x+0}{2}}{(x+0)x}\times 1+\frac{\cos x}{(x+0) \cdot x}$
$=\frac{-x \sin x}{x^{2}}-\frac{\cos x}{x^{2}}$
$=\frac{-x \sin x-\cos x}{x^{2}}$
Question 33
sin x+cos x
Sol :
माना f(x)=sin x+cos x
$f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (x+h)+\cos (x+h)-\sin x-\cos x}{h}$
$=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin x}{h}+\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h}$
$=\lim _{h \rightarrow 0} \frac{2 \sin \frac{x+4-x}{2} \cos \frac{x+2+x}{2}}{h}+\lim_{h\rightarrow 0}\dfrac{-2\sin \frac{x+h+x}{2}\sin\frac{x+h-x}{2}}{h}$
$=\lim_{h \rightarrow 0} \frac{ \sin \frac{h}{2}}{\frac{h}{2}}\lim _{h \rightarrow 0} \cos \frac{2 x+h}{2}-\lim_{h \rightarrow 0} \sin \frac{2 x+h}{2}\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}}$
$=1\times \cos \frac{2 x+0}{2}-\sin \frac{2 x+0}{2}\times 1$
=cos x-sin x
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