Another question I'm confused with. Question is:
Find the slope of the tangent to the function f(x)=2^x^2+3x when x=3.
First I start with the chain rule:
f'(x)=2^(x^2+3x) ln2 (2x+3)
Then I input the value of x=3 into the equation.
f'(3)=2^(3^2+3(3)) ln2 (2(3)+3)
f'(3)=2^(3^2+3(3)) ln2 (9)
This is where I get stuck, if I complete the function I am left with:
f'(3)= 1,635,229.37. Obviously this is not the correct answer.
I'm stuck, again.