What is the equation of the line tangent to the curve #y=(x^2-35)^7# at x=6?

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Answer 1 · 2015-05-25T19:00:49

The tangent line in a point to a function, knowing its ascissa is:

#y-f(x_0)=f'(x_0)(x-x_0)#.

So:

#f(x_0)=f(6)=(36-35)^7=1#

and

#f'(x)=7(x^2-36)^6*2xrArr#

#f'(6)=7(36-35)^6*2*6=7*1*12=84#.

The tangent line is:

#y-1=84(x-6)rArry=84x-503#.

Answer 2 · 2015-05-25T19:07:20

Try this:
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