How do you find the derivative with respect to x of #f(x)=x^2# and use it to find the equation of the tangent line to #y=x^2# at x=2?

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Answer 1 · 2017-02-18T19:05:18

#y=4x-4#

#f'(x)=2x#

Plug in 2 into this to find the slope:

#f'(2)=(2)(2)=4#

Now plug in x=2 into the first equation to get a point which will be used to find the equation.

#f(2)=2^2=4#

Now we have the point (2,4) and the slope which is 4. We can plug it into:

#(y-y_1)=m(x-x_1)#

#x_1=2#, #y_1=4#, #m=4#

#y=4=4(x-2)#

Now simplify:

#y-4=4x-8#

#y=4x-4#

We can verify the result graphically:
graph{ (y-x^2)(y-4x+4)=0 [-5, 5, -2, 12] }