Question

Find dy/dx if y=e^x/tanx​

Answer 1

Answer:

dy/dx = e^xtanx+e^xsec^2x

= e^x(tanx+sec^2x)

Answer 2

$\dfrac{dy}{dx}=e^{x}(-cosec^{2}x+cotx)$

Given:

$y=\dfrac{e^{x}}{tanx}$

To find:

$\dfrac{dy}{dx}$

Concept:

• $tanx\:cotx=1$

• $\dfrac{d}{dx}(e^{x})=e^{x}$

• $\dfrac{d}{dx}(cotx)=-cosec^{2}x$

• $\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ where u, v are functions of x

Step-by-step explanation:

Here, $y=\dfrac{e^{x}}{tanx}$

$\Rightarrow y=e^{x}cotx$

Differentiating both sides with respect to x, we get

$\dfrac{dy}{dx}=\dfrac{d}{dx}(e^{x}cotx)$

$=e^{x}\dfrac{d}{dx}(cotx)+cotx\dfrac{d}{dx}(e^{x})$

$=e^{x}(-cosec^{2}x)+e^{x}cotx$

$=e^{x}(-cosec^{2}x+cotx)$

This is the required derivative.

#SPJ3