Question

d/dx (3x)² + sinx find derivative​

Answer 1

Step-by-step explanation:

3x²d²y/dx² + 6xdy/DX + cosx

Answer 2

Solution:

Given Function:

$\rm\longrightarrow f(x)=3x^{2}+\sin(x)$

Differentiating both sides wrt x, we get:

$\rm\longrightarrow f'(x)=\dfrac{d}{dx}[3x^{2}+\sin(x)]$

$\rm\longrightarrow f'(x)=\dfrac{d}{dx}[3x^{2}]+\dfrac{d}{dx}[\sin(x)]$

We know that:

$\bigstar\ \underline{\boxed{\rm\dfrac{d}{dx}(C\times f)=C\times\dfrac{d}{dx}(f)}}$

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

$\bigstar\ \underline{\boxed{\rm\dfrac{d}{dx}[\sin(x)]=\cos(x)}}$

Using this result, we get:

$\rm\longrightarrow f'(x)=3\dfrac{d}{dx}[x^{2}]+\dfrac{d}{dx}[\sin(x)]$

$\rm\longrightarrow f'(x)=3\times 2x^{2-1}+\cos(x)$

$\rm\longrightarrow f'(x)=6x+\cos(x)$

Which is our required answer.

Learn More:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$