Question

Find te derivative of $\tt\: \frac{x}{1+tan}$

Answer 1

Given :

• $\sf{\dfrac{x}{x+tanx}}$

To Find :

• Find derivative of the given term

Solution :

• Since the given function was in fraction we can use Quotient rule.

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ${\displaystyle{\sf \frac{d}{dx}\[\left(\frac{f(x)}{g(x)}\right)\]=\frac{g(x)\frac{d}{dx}\[\left(f(x)\right)\]-f(x)\frac{d}{dx}\[\left(g(x)\right)\]}{\[\left(g(x)\right)^2\]}}}$

if $\sf{g(x) \neq 0}$

Let $\sf{f(x) =u}$ and $\sf{g(x) =v}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎${\displaystyle{\sf \frac{d}{dx}\[\left(\frac{u}{v}\right)\]=\frac{v\frac{d}{dx}\[\left(u\right)\]-u\frac{d}{dx}\[\left(v\right)\]}{\[\left(v\right)^2\]}}}$

Now substitute x in place of u and 1 + tan in place of v.

${\displaystyle{\sf \frac{d}{dx}\[\left(\frac{x}{1+tanx}\right)\]=\frac{(1+tanx)(\frac{d}{dx}\[\left(x\right)\])-(x)(\frac{d}{dx}(1+tanx))}{\[\left(1+tanx\right)^2\]}}}$

${\displaystyle{\sf \frac{d}{dx}\[\left(\frac{x}{1+tanx}\right)\]=\frac{(1+tanx)(1)-(x)(0+sec^2x)}{\[\left(1+tanx\right)^2\]}}}$

${\displaystyle{\sf \frac{d}{dx}\[\left(\frac{x}{1+tanx}\right)\]=\frac{(1+tanx)-(x sec^2x)}{\[\left(1+tanx\right)^2\]}}}$

Required Answer :

$\boxed{{\displaystyle{\sf \frac{d}{dx}\[\left(\frac{x}{1+tanx}\right)\]=\frac{(1+tanx)-(x sec^2x)}{\[\left(1+tanx\right)^2\]}}}}$