Question

find dy/dx. y =x power 3-3cosx+2tanx​

Answer 1

ANSWER:

Given:

• y = x³ - 3cosx + 2tanx

To Find:

• dy/dx

Solution:

We are given that,

$:\implies y=x^3-3\cos x+2\tan x$

Multiplying both sides by (d/dx),

$:\implies\dfrac{dy}{dx}=\dfrac{d}{dx}(x^3-3\cos x+2\tan x)$

So,

$:\implies\dfrac{dy}{dx}=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3\cos x)+\dfrac{d}{dx}(2\tan x)$

We know that,

$:\hookrightarrow\dfrac{d}{dx}x^n=nx^{n-1}$,

$:\hookrightarrow\dfrac{d}{dx}ax=a\dfrac{d}{dx}(x)\:\:(\text{a is constant})$

So,

$:\implies\dfrac{dy}{dx}=\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3\cos x)+\dfrac{d}{dx}(2\tan x)$

Hence,

$:\implies\dfrac{dy}{dx}=3x^{3-1}-3\dfrac{d}{dx}(\cos x)+2\dfrac{d}{dx}(\tan x)$

$:\implies\dfrac{dy}{dx}=3x^2-3\dfrac{d}{dx}(\cos x)+2\dfrac{d}{dx}(\tan x)$

We know that,

$:\hookrightarrow\dfrac{d}{dx}\cos x=-\sin x$,

$:\hookrightarrow\dfrac{d}{dx}\tan x=\sec^2x$

So,

$:\implies\dfrac{dy}{dx}=3x^2-3\dfrac{d}{dx}(\cos x)+2\dfrac{d}{dx}(\tan x)$

Hence,

$:\implies\dfrac{dy}{dx}=3x^2-3(-\sin x)+2(\sec^2x)$

So,

$:\implies\bf\dfrac{dy}{dx}=3x^2+3sin x+2sec^2x$

Formulae Used:

$\:\:\:\bullet\:\:\:\dfrac{d}{dx}x^n=nx^{n-1}$,

$\:\:\:\bullet\:\:\:\dfrac{d}{dx}ax=a\dfrac{d}{dx}(x)\:\:(\text{a is constant})$

$\:\:\:\bullet\:\:\:\dfrac{d}{dx}\cos x=-\sin x$,

$\:\:\:\bullet\:\:\:\dfrac{d}{dx}\tan x=\sec^2x$