if $sin x = e^{y},0 < x < pi$, what is $\frac{dy}{dx}$ in terms of $x$ \na $-\tan x$ \nb $-cot x$ \nc $cot…

if $sin x = e^{y},0 < x < pi$, what is $\frac{dy}{dx}$ in terms of $x$ \na $-\tan x$ \nb $-cot x$ \nc $cot x$ \nd $\tan x$ \ne $csc x$

if $sin x = e^{y},0 < x < pi$, what is $\frac{dy}{dx}$ in terms of $x$ \na $-\tan x$ \nb $-cot x$ \nc $cot x$ \nd $\tan x$ \ne $csc x$

Answer

Explanation:

Step1: Differentiate both sides

Differentiate $\sin x = e^{y}$ with respect to $x$. The derivative of $\sin x$ with respect to $x$ is $\cos x$, and by the chain - rule, the derivative of $e^{y}$ with respect to $x$ is $e^{y}\frac{dy}{dx}$. So we have $\cos x=e^{y}\frac{dy}{dx}$.

Step2: Solve for $\frac{dy}{dx}$

Since $\sin x = e^{y}$, substitute $e^{y}$ with $\sin x$ in the equation $\cos x=e^{y}\frac{dy}{dx}$. Then $\frac{dy}{dx}=\frac{\cos x}{\sin x}$.

Step3: Simplify the expression

We know that $\frac{\cos x}{\sin x}=\cot x$.

Answer:

C. $\cot x$