Question

differentiate the following w.r.t.x sec[tan (x^4+4)]

help me out guyz plz​

Answer 1

Given:

A function-

$\bf{sec(tan(x^{4}+4))}$

To Find:

• Differentiation of given function w.r.t. to x

Solution:

We know that,

$\leadsto\bf{\dfrac{d}{dx}(sec\:x)=sec\:x\:tan\:x}$

$\leadsto\bf{\dfrac{d}{dx}(tan\:x)=sec^{2}x}$

$\leadsto\bf{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

$\leadsto\bf{\dfrac{d}{dx}(constant)=0}$

Now,

On differentiating given function w.r.t. x, we get

$\sf{\dfrac{d}{dx}(sec(tan(x^{4}+4)))}$

$\sf{sec(tan(x^{4}+4))tan(tan(x^{4}+4))\dfrac{d}{dx}(tan(x^{4}+4))}$

$\sf{sec(tan(x^{4}+4))tan(tan(x^{4}+4))sec^{2}(x^{4}+4)\dfrac{d}{dx}(x^{4}+4)}$

$\sf{sec(tan(x^{4}+4))tan(tan(x^{4}+4))sec^{2}(x^{4}+4)(4x^{3}+0)}$

$\sf\pink{sec(tan(x^{4}+4))tan(tan(x^{4}+4))sec^{2}(x^{4}+4).4x^{3}}$

Answer 2

$y = sec(tan ( {x}^{4} + 4)) \\ \\ \frac{dy}{dx} = sec(tan ( {x}^{4} + 4) )tan(tan ( {x}^{4} + 4)) \frac{d}{dx} tan ( {x}^{4} + 4) \\ \\ \frac{dy}{dx} = sec(tan ( {x}^{4} + 4))tan(tan ( {x}^{4} + 4)) {sec}^{2} ( {x}^{4} + 4) \frac{d}{dx} ( {x}^{4} + 4) \\ \\ \frac{dy}{dx} = sec(tan ( {x}^{4} + 4))tan(tan ( {x}^{4} + 4) ) {sec}^{2} ( {x}^{4} + 4)4 {x}^{3}$