Question #1f8f7

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Question #1f8f7

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Answer 1 · 2017-11-26T13:01:20

$\frac{1}{1 + x ln (3)}$ $1/(1+xln(3))$

Explanation:

We can let $u = 1 + x ln (3)$ $u=1+xln(3)$ and use the chain rule:
$\frac{d}{d x} ({log}_{3} (1 + x ln (3))) = \frac{d}{d u} ({log}_{3} (u)) \frac{d}{d x} (1 + x ln (3))$ $d/dx(log_3(1+xln(3)))=d/(du)(log_3(u))d/dx(1+xln(3))$

Because of the rule $\frac{d}{d x} ({log}_{a} (x)) = \frac{1}{x ln (a)}$ $d/dx(log_a(x))=1/(xln(a))$ , we know:
$= \frac{1}{u ln (3)} \cdot \frac{d}{d x} (1 + x ln (3))$ $=1/(u\ln(3))*d/dx(1+xln(3))$

Since $ln (3)$ $ln(3)$ is just a constant, we can solve the other derivative too:
$=1/(ucancelln(3))*cancelln(3)=1/u$

If we resubstitute, we get our final answer:
$1/u=1/(1+xln(3))$

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