Question

d by DX of x power 4 - x power 3 + x square + ​

Answer 1

We know

$\boxed{\sf \dfrac{d(x^n)}{dx}=nx^{n+1}}$

$\\ \sf\longmapsto \dfrac{d(x^4-x^3+x^2)}{dx}$

$\\ \sf\longmapsto 4x^{4+1}-3x^{3+1}+2x^{2+1}$

$\\ \sf\longmapsto 4x^5-3x^4+2x^3$

• If we take x^3 common

$\\ \sf\longmapsto x^3(4x^2+x+2)$

Note down:-

$\boxed{\sf \dfrac{d(au)}{dx}=a\dfrac{du}{dx}}$

$\boxed{\sf \dfrac{du}{dt}=\dfrac{du}{dx}.\dfrac{dx}{dt}}$