Question

http://lhh.tutor.com/SharedSessionFiles/a93f4189-1d99-48c9-8452-1c7166aff076_pic.png

Help please?

Answer 1

Power formula can be used for differentiation \[\frac{d~x^n}{dx}=nx^{n-1}\]

Answer 2

then substitute 3 in place of x

Answer 3

we can re-write the function as \[f(x)=(5+x^2)^{1/2}\]

Answer 4

then apply the power formula

Answer 5

So I find the derivative of square root of 5+x^2,
x/square root of 5+x^2

Answer 6

if we let \(u=(5+x^2)^{1/2}\) the \(du=(1/2)(5+x^2)^{-1/2}2x~dx\)

Answer 7

yup you're right...

Answer 8

hey is that an inverse function not the derivative of the function f(x)?

Answer 9

we have to get the inverse function of f(x) before getting the 1st derivative...

Answer 10

How do I get the inverse?

Answer 11

if we let y=f(x) our equation will be \[y=\sqrt{5+x^2}\]

Answer 12

and so \[y^2=5+x^2\]\[y^2-5=5-5+x^2\]\[y^2-5=x^2\]\[\sqrt{y^2-5}=x\]and \(f^{-1}(x)=x\)

Answer 13

this is the equation we need to get the 1st derivative...

Answer 14

almost the same as before, we have just inverse the function but same calculus formula applied

Answer 15

You know that you can compute the derivative of \(f^{-1}(x)\) without having to compute the inverse...right?

Answer 16

by just replacing position of \(x^2\) and 5 and proceed to subtraction?

Answer 17

Take \(g(x)=f^{-1}(x)\). We then know that \(f(g(x))=x\). Differentiating both sides by chain rule tells us that \[f^{\prime}(g(x))\cdot g^{\prime}(x) = 1 \implies g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}.\]
Rewriting this in terms of \(f^{-1}(x)\) yields
\[(f^{-1})^{\prime}(x) = \frac{1}{f^{\prime}(f^{-1}(x))}\]

For this problem, we have \(f(x) = \sqrt{5+x^2}\) and note that \(f(2)=3\implies f^{-1}(3)=2\).

Therefore \(\large\displaystyle (f^{-1})^{\prime}(3) = \frac{1}{f^{\prime}(f^{-1}(3))} = \frac{1}{f^{\prime}(2)} = \ldots\)

Does this make sense? :-)

Answer 18

got your point... :)

Answer 19

It makes sense.

Answer 20

thanks @Orion1213 and @ChristopherToni