2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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