I am struggling with an Implicit Differentiation question which is as follows:

$z = (7x^4)*\ln(x)4$ where $z$ and $x$ are functions of $t$. $\frac{dx}{dt} = 4$ when $x = e$. Calculate $\frac{dz}{dt}$.

What I have tried so far is finding $\frac{dz}{dx}$ which I believe is $(7x^3) + (28x^3)*\ln(x)$ which I have used the chain rule to get this answer for $\frac{dz}{dx}.$

This is the part where I have tried several things such as multiplying $\frac{dz}{dx}$ by $\frac{dx}{dt}$ but I'm not sure how that works and I have tried replacing the $x$'s with $e$'s and equation $\frac{dx}{dt}$ to $\frac{dz}{dx}$ but I am really unsure what I'm supposed to be doing in this question and nothing appears right.

Any help would be greatly appreciated.

Thank you


you are almost there. what you need is $$\frac{dz}{dt} = \frac{dz}{dx}\frac{dx}{dt} = \left(7x^3+28x^3\ln x\right)\big|_{x = e} \frac{dx}{dt} = (7e^3 + 28e^3)4 = 140e^3$$

  • $\begingroup$ Thanks a lot I can't believe it. I'm kicking myself that it was that straightforward. Thank you very much. $\endgroup$ – user241451 May 18 '15 at 6:10
  • $\begingroup$ @Michael, you are welcome. overthinking can sometime cause trouble. $\endgroup$ – abel May 18 '15 at 6:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy