Calculate the slope of the tangent line at the point (2,ln(5)).

Question

Consider the Function f(x)=ln(x² )+1.

 

Answer 1

There's something wrong with the way you wrote the function.

 

(2, ln 5) does not satisfy the equation f(x) = ln(x²) + 1.

 

But if you meant f(x) = ln(x² + 1), then that works, so let's assume that's it.

 

So get the derivative...

 

By the chain rule, d/dx ln(x² + 1) = 1 / (x² + 1) * d/dx (x² + 1) = 1 / (x² + 1) * 2x = 2x / (x² + 1)

 

The slope is the value of the derivative at the point you're interested in.  x = 2, so the slope is 2*2 / (2² + 1) = 4/5.