## ksaimouli 2 years ago derivative

1. ksaimouli

|dw:1353885456771:dw|

2. ksaimouli

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

3. ksaimouli

@ivanmlerner

4. ivanmlerner

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

5. ivanmlerner

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

6. ksaimouli

|dw:1353887083835:dw|

7. ivanmlerner

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

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

9. ksaimouli

|dw:1353888159532:dw|

10. ksaimouli

|dw:1353888193965:dw|

11. ivanmlerner

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$