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

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Answer 1 · 2015-12-06T16:02:06

#y=84x-503#

Find the point the tangent line will intersect.

#f(6)=(6^2-35)^7=1#

The point is #(6,1)#.

Now, to find the slope of the tangent line, find #f'(6)#.

Finding #f'(x)# will require the chain rule:

#f'(x)=7(x^2-35)^6*d/dx[x^2-35]#

#f'(x)=7(x^2-35)^6(2x)#

#f'(x)=14x(x^2-35)^6#

#f'(6)=14(6)(6^2-35)^6=84#

Write the equation of the tangent line in point-slope form:

#y-1=84(x-6)#

In slope-intercept form:

#y=84x-503#