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

  • Thread starter esoteric deviance
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In summary, to find the value of the derivative of the inverse of the function, one should find the derivative of the original function and then use it to calculate the derivative of the inverse function. This can be done by swapping the x and y variables and then using the chain rule to differentiate.
  • #1
esoteric deviance
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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...
 
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  • #2
You just need to compute [itex]\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} [/itex]

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.
 
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  • #3
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) = [itex]f^{-1} (x)[/itex] 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) = [itex]f^{-1}[/itex]' 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 [itex]4^{\sqrt{x}}[/itex]
First look up d/dt [itex]a^{t}[/itex] in a derivative table.
Then apply chain rule.

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

Show us what that gives you or if you get the answer in the back of the book.
 
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  • #4
In general though let [itex]y=f^{-1}(x)[/itex], apply the function f to both sides to obtain [itex]f(y)=x[/itex], differentiate with respect to x to obtain:
[tex]
f'(y)\frac{dy}{dx}=1
[/tex]
Divide by f'(y) to obtain:
[tex]
\frac{dy}{dx}=\frac{1}{f(y)}=\frac{1}{f(f^{-1}(x))}
[/tex]
With your problem, swap f and the inverse of f around, so [itex]y=f(x)[/itex] and [itex]x=f^{-1}(y)[/itex]
 
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Related to Given f(x) = 4^√x - x + 3, find (f^-1)'(3).

1. What is the function f(x) given in the question?

The function is given as f(x) = 4^√x - x + 3.

2. What does (f^-1)'(3) represent?

(f^-1)'(3) represents the derivative of the inverse of the function f(x) at the point x = 3.

3. How do you find the inverse of f(x)?

To find the inverse of f(x), we switch the x and y variables and solve for y. This will give us the inverse function f^-1(x).

4. What is the process for finding the derivative of the inverse function at a given point?

The process for finding the derivative of the inverse function at a given point is to use the formula (f^-1)'(x) = 1/f'(f^-1(x)). We substitute the given point into the derivative of the original function, f'(x), and then evaluate the inverse function f^-1(x) at that point.

5. What is the solution to (f^-1)'(3) for the given function?

Using the formula (f^-1)'(x) = 1/f'(f^-1(x)), we first find the value of f^-1(3) by plugging in x = 3 into the inverse function. Then, we find the value of f'(x) by taking the derivative of the original function f(x). Finally, we substitute these values into the formula to get the solution for (f^-1)'(3).

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