Question

y=5(x^2 +3x + 2)^3 find dy/dx​

Answer 1

Answer:

Y=log(5x^2+3x+2)....given

therefore , diff^n b.s w.r.x.

dy/dx=d/dxl[og(5x^2+3x+2)]

        By chain rule,

        =(1/5x^2+3x+2)*d/dx(5x^2+3x+2)

       =[10x+3/5x^2+3x+2]

Step-by-step explanation:

Answer 2

Given :

• y = 5(x² + 3x + 2)³

To Find :

• $\sf{\dfrac{dy}{dx}}$

Solution :

Let x² + 3x + 2 = u. Then y = 5u³

• Differentiating y with respect to u ;

$\\ : \implies \bf \: \frac{dy}{du} = \frac{d}{du} (5u {}^{3} )$

• By applying , $\bf{\dfrac{d}{dx}(x^n)=\dfrac{d}{dx}(nx^{n-1})}$. We get ;

$\\ : \implies \bf \: \frac{dy}{du} = (15u {}^{2} ) \sf \: ..........(1)$

Now , differentiating u with respect to x ;

$\\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} ( {x}^{2} + 3x + 2)$

$\\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} ( {x}^{2} ) + \frac{d}{dx} (3x) + \frac{d}{dx} (3)$

• By applying , $\bf{\dfrac{d}{dx}(x^n)=\dfrac{d}{dx}(nx^{n-1})}$. We get ;

$\\ : \implies \bf \: \frac{du}{dx} = \frac{d}{dx} (2 {x}^{2 - 1} ) + \frac{d}{dx} (3 \times 1 {x}^{1 - 1} ) + \frac{d}{dx} (2)$

$\\ : \implies \bf \frac{du}{dx} = 2x + 3 \times 1 {x}^{0} + \frac{d}{dx} (2)$

• We know that a⁰ = 1. ;

$\\ : \implies \bf \: \frac{du}{dx} = 2x + 3(1) + \frac{d}{dx}(2)$

$\\ : \implies \bf \: \frac{du}{dx} = 2x + 3 + \frac{d}{dx} (2)$

• Differentitation of a Constant is "zero".

$\\ : \implies \bf \: \frac{du}{dx} = 2x + 3 + 0$

$\\ : \implies \bf \frac{du}{dx} = 2x + 3 \sf \: ...........(2)$

• Now by applying chain rule i.e ,

$\\ : \implies \bf \: \frac{dy}{du}.\frac{du}{dx} = \frac{dy}{dx}$

• From eq(1) & eq(2) . We get ;

$\\ : \implies \bf \: \frac{dy}{dx} = 15 {u}^{2} \times (2x + 3)$

• Since u = x² + 3x + 2 ;

$\\ : \implies {\underline{\boxed{\bf \: \frac{dy}{dx} = 15( {x}^{2} + 3x + 2) {}^{2} \times 2x + 3}}}$