Exercise 11.1
Question 9
Question 10
Sol :
Let y=\sin \sqrt{1+x^{2}}
Differentiating with respect to x
\frac{d y}{dx}=\frac{d(\sin \sqrt{1+x^{2}})}{d(\sqrt{1+x^{2}})} \times \frac{d(\sqrt{1+x^{2}})}{d\left(1+x^{2}\right)} \times \frac{d\left(1+x^{2}\right)}{dx}
=\cos \sqrt{1+x^{2}} \cdot \frac{1}{2 \sqrt{1+x^{2}}} \times 2 x
=\frac{x}{\sqrt{1+x^{2}}} \cos \sqrt{1+x^{2}}
\frac{d y}{dx}=\frac{d(\sin \sqrt{1+x^{2}})}{d(\sqrt{1+x^{2}})} \times \frac{d(\sqrt{1+x^{2}})}{d\left(1+x^{2}\right)} \times \frac{d\left(1+x^{2}\right)}{dx}
=\cos \sqrt{1+x^{2}} \cdot \frac{1}{2 \sqrt{1+x^{2}}} \times 2 x
=\frac{x}{\sqrt{1+x^{2}}} \cos \sqrt{1+x^{2}}
Question 11
Sol :
Let y=\sqrt{\tan 2 x}
Differentiating with respect to x
\frac{d y}{d x}=\frac{d(\sqrt{\tan 2 x})}{d(\tan 2 x)} \times \frac{d(\tan 2 x)}{d(2 x)} \times \frac{d(2 x)}{d x}
=\frac{1}{2 \sqrt{\tan 2 x}} \times \sec ^{2} 2 x \cdot 2
=\frac{\sec ^{2} 2 x}{\sqrt{\tan 2 x}}
\frac{d y}{d x}=\frac{d(\sqrt{\tan 2 x})}{d(\tan 2 x)} \times \frac{d(\tan 2 x)}{d(2 x)} \times \frac{d(2 x)}{d x}
=\frac{1}{2 \sqrt{\tan 2 x}} \times \sec ^{2} 2 x \cdot 2
=\frac{\sec ^{2} 2 x}{\sqrt{\tan 2 x}}
Question 12
Sol :
Let y=\sqrt{\sin x^{2}}
Differentiating with respect to x
\frac{d y}{d x}=\frac{d(\sqrt{\sin x^{2}})}{d\left(\sin x^{2}\right)}-\frac{d\left(\sin x^{2}\right)}{d\left(x^{2}\right)} \times \frac{d\left(x^{2}\right)}{d x}
=\frac{1}{2 \sqrt{\sin x^{2}}} \times \cos x^{2} \times 2 x
=\frac{x \cos x^{2}}{\sqrt{\sin x^{2}}}
Right ✅
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