ksaimouli
  • ksaimouli
derivative
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ksaimouli
  • ksaimouli
|dw:1353885456771:dw|
ksaimouli
  • ksaimouli
\[\ln(1-2x)^3\]
ksaimouli
  • ksaimouli
@ivanmlerner

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.