The derivative of sin(x^{3}) with respect to | vedclass

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Question

(C) Let $u = \sin(x^{3})$ and $v = \cos(x^{3})$.By differentiating both with respect to $x$,we get:$\frac{du}{dx} = \cos(x^{3}) \cdot 3x^{2}$$\frac{dv

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$-\cot(x^{3})$

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(C) Let $u = \sin(x^{3})$ and $v = \cos(x^{3})$.By differentiating both with respect to $x$,we get:$\frac{du}{dx} = \cos(x^{3}) \cdot 3x^{2}$$\frac{dv}{dx} = -\sin(x^{3}) \cdot 3x^{2}$Now,the derivative of $u$ with respect to $v$ is given by the chain rule:$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{3x^{2} \cos(x^{3})}{-3x^{2} \sin(x^{3})}$$\frac{du}{dv} = -\frac{\cos(x^{3})}{\sin(x^{3})} = -\cot(x^{3})$

The derivative of $\sin(x^{3})$ with respect to $\cos(x^{3})$ is

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