Here's the question you clicked on:
ksaimouli
derivative
|dw:1353885456771:dw|
You need to use the chain rule here. ln is one function, something^3 another and 1-2x another.
Do you have any ideas on what the first step is?
|dw:1353887083835:dw|
Almost, you forgot to derivate the function z^3. You did the derivative of the ln and the one inside the cubic power.
so should i take derivative (1-2x)^3 i thought i should only take derivative of inside the ()^3
|dw:1353888159532:dw|
|dw:1353888193965:dw|
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\[\frac{df(g(h(x)))}{dx}=\frac{df}{dg}\frac{dg}{dh}\frac{dh}{dx}\]\[\frac{df}{dg}=\frac{1}{g}=\frac{1}{(1-2x)^3}\]\[\frac{dg}{dh}=3h^2=3(1-2x)^2\]\[\frac{dh}{dx}=-2\]