Ronin Daniels

2023-03-25

What is the derivative of $f\left(x\right)=\mathrm{ln}\left[{x}^{9}{(x+3)}^{6}{({x}^{2}+7)}^{5}\right]$ ?

srnessgebf

Beginner2023-03-26Added 8 answers

Step 1

Use the properties of logarithms to rewrite:

$f\left(x\right)=\mathrm{ln}\left[{x}^{9}{(x+3)}^{6}{({x}^{2}+7)}^{5}\right]$

$=\mathrm{ln}\left({x}^{9}\right)+{\mathrm{ln}(x+3)}^{6}+{\mathrm{ln}({x}^{2}+7)}^{5}$

$=9\mathrm{ln}\left(x\right)+6\mathrm{ln}(x+3)+5\mathrm{ln}({x}^{2}+7)$

So,

$f\prime \left(x\right)=\frac{9}{x}+\frac{6}{x+3}+\frac{5}{{x}^{2}+7}2x$

Simplify algebraically as desired.

Use the properties of logarithms to rewrite:

$f\left(x\right)=\mathrm{ln}\left[{x}^{9}{(x+3)}^{6}{({x}^{2}+7)}^{5}\right]$

$=\mathrm{ln}\left({x}^{9}\right)+{\mathrm{ln}(x+3)}^{6}+{\mathrm{ln}({x}^{2}+7)}^{5}$

$=9\mathrm{ln}\left(x\right)+6\mathrm{ln}(x+3)+5\mathrm{ln}({x}^{2}+7)$

So,

$f\prime \left(x\right)=\frac{9}{x}+\frac{6}{x+3}+\frac{5}{{x}^{2}+7}2x$

Simplify algebraically as desired.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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