SOLUTION: Use limits to find the slope of the tangent to f(x) = x^2, at the point (3,9). Please use the differentiation from the first principles method if possible, as I don't understan

Algebra ->  Functions -> SOLUTION: Use limits to find the slope of the tangent to f(x) = x^2, at the point (3,9). Please use the differentiation from the first principles method if possible, as I don't understan      Log On


   

Question 1190590: Use limits to find the slope of the tangent to f(x) = x^2, at the point (3,9).
Please use the differentiation from the first principles method if possible, as I don't understand how it would work here.
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

f(x) = x^2
f(x+h) = (x+h)^2 ... every x is replaced with x+h
f(x+h) = x^2+2xh+h^2 ... expand with the FOIL rule

Apply the difference quotient
%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28%28x%5E2%2B2xh%2Bh%5E2%29-%28x%5E2%29%29%2Fh

%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28x%5E2%2B2xh%2Bh%5E2-x%5E2%29%2Fh

%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%282xh%2Bh%5E2%29%2Fh The x^2 terms cancel out, leaving nothing but terms with h in the numerator

%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28h%282x%2Bh%29%29%2Fh That way we can factor out h



%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28cross%28h%29%282x%2Bh%29%29%2F%28cross%28h%29%29 The h's cancel

%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+2x%2Bh
Now let h approach 0. It won't actually reach 0 itself, but only get closer and closer. This is the limit process in action. Effectively we replace h with 0 (refer to the substitution limit theorem) and we go from 2x+h to 2x+0 or simply 2x.

In short, the derivative of f(x) = x^2 is f ' (x) = 2x.
You'll learn a shortcut (if you haven't already) that's useful and that's the power rule. That will mean you can quickly differentiate any polynomial like this without having to resort to these limits. However it's good to know how all this works out.

Plug x = 3 into the derivative function to find the tangent slope
f ' (x) = 2x
f ' (3) = 2*3
f ' (3) = 6
The tangent slope is m = 6

Answer: 6

Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Actually,  you ask us,  the tutors,  re-write again specially for you what is written in tens  Calculus textbooks.

            Every  Calculus student must have and read at least one such source.

                    If you do not understand what is written there,  then you deserve a special reward . . .

            If you do not have such a textbook on your table,  you may learn the subject from the Internet.


To learn on " HOW TO differentiate from first principles ",   read from these  Internet sources


https://www.siyavula.com/read/maths/grade-12/differential-calculus/06-differential-calculus-02

https://owlcation.com/stem/How-to-Differentiate-from-First-Principles


You will find full explanations there.