Question

Find dy/dx for the y=x to the power 5+x to the power3+10

Answer 1

Given

$\sf \to \: y \: = {x}^{5} + {x}^{3} + 10$

To Find

$\sf \to \: \dfrac{dy}{dx}$

Now Put the value of y

$\sf \to \: \dfrac{d( {x}^{5} + {x}^{3} + 10) }{dx}$

Use this Properties

$\sf \to \: \dfrac{d(ax \pm bx)}{dx} = \dfrac{d(ax)}{dx} \pm \dfrac{d(bx)}{dx}$

We Get

$\sf \to \: \dfrac{d( {x}^{5}) }{dx} + \dfrac{d( {x}^{3} )}{dx} + \dfrac{d(10)}{dx}$

Formula we use

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

$\to \sf \dfrac{d(a)}{dx} = 0 \: \: \: \: \: \: \: \: \: where \: a \: is \: constant$

Now

$\sf \to \: 5 {x}^{5 - 1} + 3 {x}^{3 - 1} + 0$

$\sf \to \: 5 {x}^{4} + 3 {x}^{2}$

Answer

$\to\sf \: \dfrac{dy}{dx} = \sf \: 5 {x}^{4} + 3 {x}^{2}$