Question

what is derivative tan (4+2x)​

Answer 1

$let \: \: y = \tan(4 + 2x)$

So

Differentiating both sides with respect to x we get

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

$\frac{dy}{dx} = 2 \: {sec}^{2} (4 + 2x)$

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Answer 2

Given :

Function : y = tan(4+2x)

To Find :

dy/dx

Solution :

We have ,

$\sf\:y=\tan(4+2x)$

let 4+2x= u, then

$\sf\:y=\tan(u)$

Now Differentiate with respect to x , by chain rule

$\sf\dfrac{dy}{dx}=\dfrac{d(tan\:u)}{du}\times\dfrac{du}{dx}$

We know that , $\sf\dfrac{d(\tan\:x) }{dx} =\sec^2\:x$

$\sf\implies\dfrac{dy}{dx}=\sec^2u\times\dfrac{du}{dx}$

$\sf\implies\dfrac{dy}{dx}=\sec^2(4+2x)\times\dfrac{du}{dx}..(1)$

And ,we have

u = 4+2x

Now Differentiate with respect to x

$\sf\dfrac{du}{dx}=0+2$

$\sf\dfrac{du}{dx}=2$

Now put the value of du/dx in Equation (1),then

${\purple{\boxed{\large{\bold{\dfrac{dy}{dx}=\sec^2(4+2x)\times2}}}}}$

$\rule{200}2$

${\underline{\sf{Formula's}}}$

$1)\sf\:\frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2)\sf\:\frac{d(constant)}{dx} = 0$

$3) \sf\dfrac{d(\cos\:x)}{dx} =-\sin\:x$

$\sf4) \dfrac{d(\sin\:x) }{dx} =\cos\:x$

${\red{\boxed{\large{\bold{Composite\: Function (Chain\:Rule)}}}}}$

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$