Diffrentiate the following w.r.t.x x+ x - Vidyadip

Question

Diffrentiate the following w.r.t.x

$\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$

✓

Answer

Let $y =\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$

Differentiating w.r.t. x, we get

$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}] \\ & =\frac{d}{d x}(\cos x)^{\frac{1}{2}}+\frac{d}{d x}(\cos \sqrt{x})^{\frac{1}{2}} \\ & =\frac{1}{2}(\cos x)^{-\frac{1}{2}} \cdot \frac{d}{d x}(\cos x)+\frac{1}{2}(\cos \sqrt{x})^{-\frac{1}{2}} \cdot \frac{d}{d x}(\cos \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\cos x}} \cdot(-\sin x)+\frac{1}{2 \sqrt{\cos \sqrt{x}}} \times(-\sin \sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{\sin \sqrt{x}}{2 \sqrt{\cos \sqrt{x}}} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{\sin \sqrt{x}}{4 \sqrt{x} \sqrt{\cos \sqrt{x}}} .\end{aligned}$

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