3 years ago

It is exponent laws,I don't get it so can you go step by step plz

Mathematics

1 answer:

ipn [44]3 years ago

5 0

I hope this helps you

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adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Hence, the required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

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2 years ago

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Artyom0805 [142]

Answer:

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Step-by-step explanation:

P.S. im not doing your hw

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2 years ago

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a_sh-v [17]

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