Question

Find the derivative of x+cosx/tanx with respect to x

Answer 1

The derivative of the given expression is $\frac{tanx-tanxsinx-xsec^2x-secx}{tan^2x}$

Therefore $\frac{dy}{dx}=\frac{tanx-tanxsinx-xsec^2x-secx}{tan^2x}$

Step by step explanation :

Given that $\frac{x+cosx}{tanx}$

To find the $\frac{dy}{dx}$ for the given expression  :

Let y=$\frac{x+cosx}{tanx}$

Differentiating with respect to x

$\frac{dy}{dx}=\frac{tanx(1+(-sinx))-(x+cosx)(sec^2x)}{tan^2x}$ ( by using the formula $\frac{u}{v}=\frac{vdu-udv}{v^2}$ here u=x+cosx and v=tanx )

$=\frac{tanx(1-sinx)-(x+cosx)(sec^2x)}{tan^2x}$

$=\frac{tanx-tanxsinx-xsec^2x-cosxsec^2x}{tan^2x}$

$=\frac{tanx-tanxsinx-xsec^2x-(\frac{1}{secx})sec^2x}{tan^2x}$

$=\frac{tanx-tanxsinx-xsec^2x-secx}{tan^2x}$

Therefore $\frac{dy}{dx}=\frac{tanx-tanxsinx-xsec^2x-secx}{tan^2x}$

The derivative of the given expression is $\frac{tanx-tanxsinx-xsec^2x-secx}{tan^2x}$

Answer 2

Step-by-step explanation:

above answer will help u