What is the equation of the tangent line of #f(x)=ln(x^3-7)# at #x=2#?

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Answer 1 · 2016-06-08T00:06:42

#y=12x-24#

First, find the point of tangency by finding the function value at #x=2#.

#f(2)=ln((2)^3-7)=ln(1)=0#

The tangent line will go through the point #(2,0)#.

To find the slope of the tangent line, find the value of the function's derivative at #x=2#.

To differentiate the function, we will use the chain rule.

Since #d/dxln(x)=1/x#, we see that #d/dxln(g(x))=1/g(x)*g'(x)#.

Therefore

#f'(x)=1/(x^3-7)*d/dx(x^3-7)=(3x^2)/(x^3-7)#

The slope of the tangent line at #x=2# is

#f'(2)=(3(2)^2)/((2)^3-7)=(3(4))/(8-7)=12/1=12#

The equation of the line with slope #12# passing through the point #(2,0)# is

#y-0=12(x-2)" "=>" "y=12x-24#