Question

$\bold \red{\small{Find \: the \: derivative \: of}}$
$\bold \red{f(x) = 2x {}^{5} +3x}$
•Don't spam
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {2x}^{5} + 3x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}f(x) = \dfrac{d}{dx}( {2x}^{5} + 3x)$

We know

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v \: }}}$

So, using this, we get

$\rm :\longmapsto\:f'(x) = \dfrac{d}{dx} {2x}^{5} + \dfrac{d}{dx}3x$

We know,

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x) \: }}}$

So, using this, we get

$\rm :\longmapsto\:f'(x) =2 \dfrac{d}{dx} {x}^{5} + 3\dfrac{d}{dx}x$

Now, we know that

$\red{\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}}$

So, using this, we get

$\rm :\longmapsto\:f'(x) =2 \times {5x}^{5 - 1} + 3 \times 1$

$\rm :\longmapsto\:f'(x) ={10x}^{4} + 3$

Hence,

$\purple{\rm \implies\:\boxed{ \bf{ \: \dfrac{d}{dx}( {2x}^{5} + 3x) ={10x}^{4} + 3 \: }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$

Answer 2

Given : $\rm f(x) = 2x^{5} +3x$

To find : Derivative of the given function

Solution :

To solve this problem, we should know some basic formulas for derivatives.

• $\boxed{ \rm\dfrac{d}{dx}(x^{n} ) =n {x}^{n - 1} }$

• $\boxed{\rm \dfrac{d}{dx}k.f(x ) =k.\dfrac{d}{dx}f(x ) }$

• $\boxed{ \rm\dfrac{d}{dx}(v\pm u)= \frac{d}{dx}(v )\pm \dfrac{d}{dx}(u) }$

Now let's solve the given problem!!

$\rm\implies f(x) = 2x^{5} +3x$

On differentiating both sides w.r.t. x, we get :

$\rm \implies f'(x) = \dfrac{d}{dx}(2x^{5} +3x)$

$\rm \implies f'(x) = \dfrac{d}{dx}(2x^{5} +3x)$

$\rm \implies f'(x) = \dfrac{d}{dx}(2x^{5}) + \dfrac{d}{dx} (3x)$

$\rm \implies f'(x) = 2.\dfrac{d}{dx}(x^{5}) + \dfrac{d}{dx} (3x)$

$\rm \implies f'(x) = 2(5)(x^{5 - 1}) + (3x {}^{1 - 1} )$

$\rm \implies f'(x) = 10(x^{4}) + (3x {}^{0} )$

$\rm \implies f'(x) = 10(x^{4}) + (3.1 )$

$\rm \implies f'(x) = 10x^{4}+3$

Hence the required answer is,

$\pink{ \boxed{\rm \dfrac{d}{dx} f(x) =10 {x}^{4} + 3}}$

‎ ‎ ‎

Learn More :

Find the differentiation or sin x

https://brainly.in/question/45317798

Prove the chain rule of differentiation

https://brainly.in/question/45318358