Math, asked by MorningstarSayoo, 3 months ago

y=5(x^2 +3x + 2)^3 find dy/dx​

Answers

Answered by Anonymous
3

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:

Answered by Mysterioushine
55

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}}}

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