Given f(x) = 4^√x - x + 3, find (f^-1)'(3).

[esoteric deviance]

Homework Statement

Find the value of the derivative of the inverse of the function.

Given that f(x) = (4^√x) - x + 3, find (f^-1)'(3).

Homework Equations

The Attempt at a Solution

I know that to find the inverse derivative, one should switch the x and y's and then take the derivative, but that 4^√x is kind of throwing me off...

 

[Pyrrhus]

You just need to compute $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$

You specify z= (4^√x), and then calculate dz/dx. You can do that by applying natural log as ln z = √x ln 4, take derivative, and then go back to compute the original dz/dx.

You can do this because if h(t) = u(t) + w(t) and you know the differential operator has the linear property, and thus h' = u' + w', where the primes mean derivatives wrt to t.

 

[nickalh]

Pyrrhus, is right, but from experience many introductory calculus students won't understand everything he's saying.


http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx"
Keep in mind that site uses f(x)=f(x) sometimes and g(x) = $f^{-1} (x)$ other times.
So context is essential for knowing if x is the horizontal, independent variable or x might better be thought of as the vertical, dependent variable. Alternately, the notation and concepts may be easier to understand if we ignore the usual interpretation of x vs. y.

To evaluate g'(3) = $f^{-1}$' we'll need to first find
g(3). We need to apply f( g(3) ) = 3 to find g(3).
So f(x) = 3.
I'll let you work out x. For now, I'll call it b.
Next step, find f'. You'll need.

d/dx $4^{\sqrt{x}}$
First look up d/dt $a^{t}$ in a derivative table.
Then apply chain rule.

I trust you can find
g'(3) = $\frac{1}{f'(b)}$
from there.

Show us what that gives you or if you get the answer in the back of the book.

 

[hunt_mat]

In general though let $y=f^{-1}(x)$, apply the function f to both sides to obtain $f(y)=x$, differentiate with respect to x to obtain:
$$f'(y)\frac{dy}{dx}=1$$
Divide by f'(y) to obtain:
$$\frac{dy}{dx}=\frac{1}{f(y)}=\frac{1}{f(f^{-1}(x))}$$
With your problem, swap f and the inverse of f around, so $y=f(x)$ and $x=f^{-1}(y)$