If y = cos(x^{circ}) and z = cos x,then frac{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y = \cos(x^{\circ})$ and $z = \cos x$. Since $x^{\circ} = \frac{\pi x}{180}$ radians,we have $y = \cos\left(\frac{\pi x}{180}\right)$. Diff

Vedclass · Accepted Answer

$\frac{\pi}{180} \sin(x^{\circ}) \operatorname{cosec} x$

Vedclass · Answer

(C) Given $y = \cos(x^{\circ})$ and $z = \cos x$. Since $x^{\circ} = \frac{\pi x}{180}$ radians,we have $y = \cos\left(\frac{\pi x}{180}\right)$. Differentiating $y$ with respect to $x$: $\frac{dy}{dx} = -\sin\left(\frac{\pi x}{180}\right) \cdot \frac{\pi}{180} = -\frac{\pi}{180} \sin(x^{\circ})$. Differentiating $z$ with respect to $x$: $\frac{dz}{dx} = -\sin x$. Now,using the chain rule: $\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{-\frac{\pi}{180} \sin(x^{\circ})}{-\sin x} = \frac{\pi}{180} \sin(x^{\circ}) \operatorname{cosec} x$.

If $y = \cos(x^{\circ})$ and $z = \cos x$,then $\frac{dy}{dz}$ is equal to

Similar Questions

Differentiate the following with respect to $x$: $e^{-x}$

If $f(x)=\cosh ^{-1}\left(\frac{1-x}{1+x}\right)$ is well defined, then $f^{\prime}(x)=$

$y=\log \left\{\left(\frac{1+x}{1-x}\right)^{1 / 4}\right\}-\frac{1}{2} \tan ^{-1}(x)$,then $\frac{d y}{d x}$ is equal to

The derivative of ${x^6} + {6^x}$ with respect to $x$ is:

If $f^{\prime}(x)=\tan^{-1}(\sec x+\tan x)$ for $-\frac{\pi}{2} < x < \frac{\pi}{2}$ and $f(0)=0$,then $f(1)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module