What is the derivative of #x=y^2#?

1 Answer
Dec 3, 2014

We can solve this problem in a few steps using Implicit Differentiation.
Step 1) Take the derivative of both sides with respect to x.

  • #(Delta)/(Deltax)(y^2)=(Delta)/(Deltax)(x)#

Step 2) To find #(Delta)/(Deltax)(y^2)# we have to use the chain rule because the variables are different.

  • Chain rule: #(Delta)/(Deltax)(u^n)= (n*u^(n-1))*(u')#

  • Plugging in our problem: #(Delta)/(Deltax)(y^2)=(2*y)*(Deltay)/(Deltax)#

Step 3) Find #(Delta)/(Deltax)(x)# with the simple power rule since the variables are the same.

  • Power rule: #(Delta)/(Deltax)(x^n)= (n*x^(n-1))#

  • Plugging in our problem: #(Delta)/(Deltax)(x)=1#

Step 4) Plugging in the values found in steps 2 and 3 back into the original equation ( #(Delta)/(Deltax)(y^2)=(Delta)/(Deltax)(x)# ) we can finally solve for #(Deltay)/(Deltax)#.

  • #(2*y)*(Deltay)/(Deltax)=1#

Divide both sides by #2y# to get #(Deltay)/(Deltax)# by itself

  • #(Deltay)/(Deltax)=1/(2*y)#

This is the solution

Notice: the chain rule and power rule are very similar, the only differences are:
-chain rule: #u!=x# "variables are different" and
-power rule: #x=x# "variables are the same"