Question

Q1.if y=(x power 2+1) (2x-1) find dy/ dx.

Answer 1

Step-by-step explanation:

We have to differentiate the following expression :-

$\rm y = ({x}^{2} + 1)(2x - 1)$

$\rm \longrightarrow \dfrac{dy}{dx} = \dfrac{d \bigg(({x}^{2} + 1)(2x - 1)\bigg)}{dx}</p><p>$

$\rm \longrightarrow \dfrac{dy}{dx} = (2x - 1) \dfrac{d \bigg({x}^{2} + 1\bigg)}{dx} + ({x}^{2} + 1) \dfrac{d \bigg(2x - 1\bigg)}{dx}</p><p>$

$\rm \longrightarrow \dfrac{dy}{dx} = (2x - 1) (2x) + ({x}^{2} + 1) 2</p><p>$

$\rm \longrightarrow \dfrac{dy}{dx} = 4 {x}^{2} - 2x + 2{x}^{2} +2$

$\rm \longrightarrow \dfrac{dy}{dx} = 6{x}^{2} - 2x +2</p><p>$

Answer :-

$\rm \dfrac{dy}{dx} = 6{x}^{2} - 2x +2 </p><p>$

Additional Information :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x

Answer 2

Answer:

Given,

$\sf \: y = ( {x}^{2} + 1)(2x - 1)$

We have to find its first derivatives.

But, firstly we have to know these :

$\pink{ \odot \: \sf \: \frac{d \: uv}{dx} = u \frac{d \: v}{dx} + v \frac{d \: u}{dx} }\\ \\ \orange{ \odot \: \: \sf \frac{d(u \pm v)}{dx} = \frac{d \: u}{dx} \pm \: \frac{d \: v}{dx} }\\ \\ \blue{ \odot \: \: \sf \: \frac{d {x}^{n} }{dx} = n {x}^{n - 1} } \\ \\ \red {\odot \sf \: \frac{d \: (constant)}{dx} = 0}$

Solution :

$\sf \: \frac{dy}{dx} = \frac{d \: ( {x}^{2} + 1)(2x - 1)}{dx} \\ \\ \sf \: \: \: \: \: \: \: = ( {x}^{2} + 1) \frac{d \: (2x - 1)}{dx} + (2x - 1) \frac{d \: ( {x}^{2} + 1) }{dx} \\ \\ \sf \: \: \: \: \: \: \: = ( {x}^{2} + 1) \{ \frac{d \: 2x}{dx} - \frac{d \:( 1)}{dx} \} + (2x - 1) \{ \frac{d \: {x}^{2} }{dx} + \frac{d \: (1)}{dx} \} \\ \\ \sf \: \: \: \: \: \: \: = ( {x}^{2} + 1)(2 - 0) + (2x - 1)(2x + 0) \\ \\ \sf \: \: \: \: \: \: \: = 2( {x}^{2} + 1) + 2x(2x - 1) \\ \\ \sf \: \: \: \: \: \: \: = 2 {x}^{2} + 2 + 4 {x}^{2} - 2 x\\ \\ \sf \: \: \: \: \: \: \: = 6 {x}^{2} - 2x + 2\\ \\ \\ \\ \: \: \: \: \: \: \: \: \green{ \boxed {\sf \therefore \frac{dy}{dx} = 6 {x}^{2} - 2x + 2}}$