-3

Find $dz/dt$ for $z=e^x \cos y$, where $x$ and $y$ are functions of $t$ defined by$$x^3+e^x-t^2-t=1\text{ and }yt^2+y^2t-t+y=0.$$

I try to use chain rule that dz/dt = (dz/dx)(dx/dt)+(dz/dy)(dy/dt), but I don't know how to find dx/dt and dy/dt in this problem.

choco_addicted
  • 8,786
  • 8
  • 24
  • 53
user133140
  • 885
  • Are you sure you mean partial differentiation? This looks a lot like regular differentiation to me. –  Mar 22 '14 at 23:30
  • Because if we can write x=g(t) and y=h(t),then it should be partial differentiation, we can easily solve it by chain rule, but for this I am not sure. – user133140 Mar 22 '14 at 23:36
  • Partial differentiation should give an answer of $0$. –  Mar 22 '14 at 23:37

1 Answers1

1

For normal differentiation, $$\dfrac{dz}{dt}=-e^x\sin y\dfrac{dy}{dt}+\cos ye^x\dfrac{dx}{dt}$$ Implicit differentation thus gives $$3x^2\dfrac{dx}{dt}+e^x\dfrac{dx}{dt}-2t-1=0\\ t^2\dfrac{dy}{dt}+2ty+y^2+2ty\dfrac{dy}{dt}-1+\dfrac{dy}{dt}=0$$ Thus, $$3x^2\dfrac{dx}{dt}+e^x\dfrac{dx}{dt}-2t-1=0\\ \implies \dfrac{dx}{dt}=\dfrac{1+2t}{3x^2+e^x}$$ And $$t^2\dfrac{dy}{dt}+2ty+y^2+2ty\dfrac{dy}{dt}-1+\dfrac{dy}{dt}=0\\ \implies \dfrac{dy}{dt}=\dfrac{1-2ty+y^2}{t^2+2ty+1}$$ Substitute: $$\boxed{\dfrac{dz}{dt}=-e^x\sin y\dfrac{1-2ty+y^2}{t^2+2ty+1}+e^x\cos y\dfrac{1+2t}{3x^2+e^x}}$$

  • I got it, thanks! I never seen this kind of question before, I think I need more practice. – user133140 Mar 22 '14 at 23:41
  • Note that your question did not ask about partial differentiation - a partial derivative of a function of several variables is its derivative with respect to one of those variables, with the others held constant, so partial differentiation would have made $\dfrac{dz}{dt}=0$. –  Mar 22 '14 at 23:43
  • Ok,I'll go over this chapter. – user133140 Mar 22 '14 at 23:46