Question

Find (f^-1)'(a). f(x)=8+x^2+tan(xpi/2), -1 11 years ago

Answer 1

\[f(x) ^{-1}=\frac{ 1 }{ f(x) }=\frac{ 1 }{ 8+x^{2} \tan \frac{ \Pi }{ 2 } x}\]
\[([f(x)] ^{-1}) \prime =\frac{ 0-1*(0+2x+\frac{\Pi }{ 2 } \sec ^{2}\frac{ \Pi }{2 } x) }{ (8+x ^{2}+\tan \frac{ \Pi }{ 2}x )^{2}}\]
\[([f(8)]^{-1}) \prime=\frac{ 2 +\frac{ \Pi } {2}\sec ^{2}(\frac{ \Pi }{ 2 }*8)}{8+8^{2}+\tan(\frac{ \Pi }{ 2}*8 ) }\]
\[([f(8)]^{-1}) \prime =\frac{ 1 }{ 36 } + \frac{ \Pi }{ 72 }\]
\[\approx.05\]

Answer 2

I had difficulty following the calculations posted above by ENG.M.A.MOTAAL, but that is perhaps that I interpreted the problem as being somewhat simpler. So please compare our approaches to make sure that I have not over-simplified everything. So..

My approach differed via the following sequence of calculations; noting that I seem to type "x" instead of using the letter "a" below when typing the differential symbol (which is a habit but should be inconsequential here).

1) The principal equation: f(x)=8+x^2+tan(xpi/2)

As I understand the syntax above, we are asked to perform the first derivative as an inverse function whilst using "a" integer (8) as the value to be derived. Yes? Therefore using the value of "8" inserted into the derivative of an inverse function, there are two ways to approach this calculation this:

First approach: This is or seems a straightforward 1/f'(8) first derivative of the equation (1) above and yields (1/16).

What I actually did to simplify this in my brain was to obtain the first derivative, f'(8), of f(x)=8+x^2+tan(xpi/2), which yield us "16" as the answer, then I simply inverted the answer to get (1/16).

A Second Approach: If you wish to derive [1/f'(8)] straight away, then you get (1/16) "unmittelbar" (immediately) as we say back home.

Last and Third Way: If you wish to obtain the inverse derivative as an equation, and then substitute for the x, then (1/f'[x]) of f(x)=8+x^2+tan(xpi/2) yields us (1/2x) as the answer, und then we substitute x=8 and get (1/16) as the answer again.

Thank you for posting this problem and my best wishes for your success. I hope that Calculus becomes a relaxed and enjoyable endeavor and I wish that I have not complicated things for you.
Best wishes and Auf Wiedersehen,
Felsen

Answer 3

Hi Felsen and every body. First of all you should know that
\[\frac{ 1 }{ f(x) \prime } \neq (\frac{ 1 }{ f(x) })\prime \]
for example : assume f(x)=x^2
then :
1.\[f(x)=x^{3}\]
2.\[f(x) \prime =3x^{2}\]
3.\[\frac{ 1 }{ f(x) \prime}=\frac{ 1 }{ 3 } x^{-2}\] substitute with x=2 then
\[\frac{ 1 }{ f(3)\prime }=\frac{1}{3}*\frac{1}{4}=\frac{1}{12}\]
4. on the other side:
\[(\frac{1}{f(x)})\prime=(\frac{1}{x^{3}})\prime =-3x^{-4}\]
and if we substitute by the same value of x which is 2 we get:
\[(\frac{1}{f(2)})\prime =-3*(2)^{-4}=\frac{-3}{16}\]
which is not equal to 1/12
the main trick of the problem is the concept that
\[(\frac{1}{f(x)})\prime \neq \frac{1}{f(x)\prime}\]
I tried to simplify it as much as I can but the major difficulty in the problem lies in the previous trick .....just concentrate a little bit you 'll get it

Answer 4

sorry my first answer is wrong but the main concept is write sorry i answered on a hurry I'll denote the right answer later

Answer 5

@ENG.M.A.MOTAAL
though the notation is ambiguous, the question is probably referring to the *inverse* function, and we use this property:

Let y= f(x) be continuous and increasing on the interval (a,b), with inverse
x = \(\phi(y) \). Then if f(x) is differentable at x= \(x_0 \in (a,b)\), with f'(x_0) \( \ne \) 0
Then
\[ \phi' (y_0) = \frac{1}{f'(x_0)} \]

using
\[ \frac{d}{dx} 8+x^2+\tan\left(\frac{\pi}{2}x\right) = 2x + \frac{\pi}{2} \sec^2\left(\frac{\pi}{2}x\right)\]
we have
\[ \phi' (y_0) = 2x_0 + \frac{\pi}{2} \sec^2\left(\frac{\pi}{2}x_0\right)\]

with \(y_0=8\) we use f(x) to find \(x_0\)
\[ 8+x^2+\tan\left(\frac{\pi}{2}x\right) = 8 \\
x^2+\tan\left(\frac{\pi}{2}x\right)=0
\]
which looks difficult to solve, except by inspection, we notice x=0 works
thus:
\[ \phi' (8) = 2\cdot 0 + \frac{\pi}{2} \sec^2\left(0\right) \\= \frac{\pi}{2} \]

Answer 6

though the notation is ambiguous, the question is probably referring to the *inverse* function, and we use this property: Let y= f(x) be continuous and increasing on the interval (a,b), with inverse x = ϕ(y). Then if f(x) is differentiable at x= x0∈(a,b), with f'(x_0) ≠ 0 Then \[ \phi' (y_0) = \frac{1}{f'(x_0)} \]
\[f'(x)= \frac{d}{dx} 8+x^2+\tan\left(\frac{\pi}{2}x\right) = 2x + \frac{\pi}{2} \sec^2\left(\frac{\pi}{2}x\right) \]
we have
\[ \phi' (y_0) =\frac{1 }{ 2x_0 + \frac{\pi}{2} \sec^2\left(\frac{\pi}{2}x_0\right) }\]
with \(y_0=8\) we use f(x) to find \(x_0\)
\[ f(x_0)=8\\8+x_0^2+\tan\left(\frac{\pi}{2}x_0\right) = 8 \\
x_0^2+\tan\left(\frac{\pi}{2}x_0\right)=0
\]
which looks difficult to solve, except by inspection, we notice x=0 works thus:
\[ \phi' (8) = \frac{1}{2\cdot 0 + \frac{\pi}{2} \sec^2\left(0\right) }\\ = \frac{2}{\pi}\\
\]