$$x\sin(4x+5y)=y\cos(x)$$ I am trying to use implicit differentiation to find dx/dy for this problem but the answer i keep getting is $$4x\cos(4x+5y)=-y\sin(x)$$ and I am stuck.

  • $\begingroup$ You need to use the product rule when you differentiate both sides. $\endgroup$ – user84413 Mar 8 '14 at 0:01

You're forgetting that $x$ is dependent on $y$ so you have to take chain rule into account. I'll do part of it. Hopefully you can see what to do from there.

$$\begin{align}\frac{d}{dy}(x\sin(4x+5y)) &\stackrel{\text{product rule}}{=} \left(\frac{d}{dy}x\right)\sin(4x+5y)+x\frac{d}{dy}\sin(4x+5y)\\ &\stackrel{\text{chain rule}}{=} \frac{dx}{dy}\sin(4x+5y) + x\cos(4x+5y)\cdot\frac{d}{dy}(4x+5y)\\ &= \frac{dx}{dy}\sin(4x+5y)+x\cos(4x+5y)\left(4\frac{dx}{dy}+5\right)\end{align}$$

Does this make sense?

  • $\begingroup$ is this just the left side of the equation $\endgroup$ – Kelli Davis Mar 8 '14 at 0:41
  • $\begingroup$ Yes it is. You should have a clearer idea of what to do with the right side, I hope. $\endgroup$ – Cameron Williams Mar 8 '14 at 0:43
  • $\begingroup$ i'm still having problems. can you break it down a little more? $\endgroup$ – Kelli Davis Mar 8 '14 at 2:46
  • $\begingroup$ What specifically are you having trouble with? $\endgroup$ – William Chang Mar 8 '14 at 5:06
  • $\begingroup$ @Mathster I used the product and chain rule on both sides. I keep getting (sin(4x+5y) + 4xcos(4x+5y)+ysinx)/cosx. but that isnt the right answer $\endgroup$ – Kelli Davis Mar 8 '14 at 5:29

$$\frac{d}{dx}\left(x\sin(4x+5y)=y\cos(x)\right)\\ \implies \sin(4x+5y)+x\frac{d}{dx}\sin(4x+5y)=\cos(x)\frac{dy}{dx}+y\frac{d}{dx}\cos(x)$$ Let us differentiate part-by-part. $$\frac{d}{dx}\sin(4x+5y)=\frac{d}{d(4x+5y)}(\sin(4x+5y))\cdot\left(20y+20x\frac{dy}{dx}\right)=\\20\cos(4x+5y)\left(y+x\frac{dy}{dx}\right)$$ Then, $$\frac{d}{dx}\cos(x)=-\sin(x)$$ Substitution gives $$\sin(4x+5y)+x\frac{d}{dx}\sin(4x+5y)=\cos(x)\frac{dy}{dx}+y\frac{d}{dx}\cos(x)\\ \implies\sin(4x+5y)+20x\cos(4x+5y)\left(y+x\frac{dy}{dx}\right)=\cos(x)\frac{dy}{dx}-y\sin(x)\\ \implies \sin(4x+5y)+20x\cos(4x+5y)y+y\sin(x)=\left(\cos(x)-20x^2\cos(4x+5y)\right)\frac{dy}{dx}\\ \implies\boxed{\dfrac{dy}{dx}=\dfrac{\sin(4x+5y)+20x\cos(4x+5y)y+y\sin(x)}{\cos(x)-20x^2\cos(4x+5y)}}$$

  • $\begingroup$ this is not the right answer $\endgroup$ – Kelli Davis Mar 8 '14 at 5:03

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.