## ksaimouli Group Title derivative one year ago one year ago

1. ksaimouli Group Title

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2. ksaimouli Group Title

$\ln(1-2x)^3$

3. ksaimouli Group Title

@ivanmlerner

4. ivanmlerner Group Title

You need to use the chain rule here. ln is one function, something^3 another and 1-2x another.

5. ivanmlerner Group Title

Do you have any ideas on what the first step is?

6. ksaimouli Group Title

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7. ivanmlerner Group Title

Almost, you forgot to derivate the function z^3. You did the derivative of the ln and the one inside the cubic power.

8. ksaimouli Group Title

so should i take derivative (1-2x)^3 i thought i should only take derivative of inside the ()^3

9. ksaimouli Group Title

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10. ksaimouli Group Title

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11. ivanmlerner Group Title

Now that is correct. Just to summarize you have the following:$f(x)=\ln(x)$$g(x)=x^3$$h(x)=1-2x$and you want$\frac{df(g(h(x)))}{dx}=\frac{df}{dg}\frac{dg}{dh}\frac{dh}{dx}$$\frac{df}{dg}=\frac{1}{g}=\frac{1}{(1-2x)^3}$$\frac{dg}{dh}=3h^2=3(1-2x)^2$$\frac{dh}{dx}=-2$