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If $$f(x) = x^2 + x + e^x\\g(x) = f^{-1}(x) $$ then use the chain rule to find $g'(1)$.

When I tried to find the inverse of $f(x)$ I couldn't as it is not a one-to-one function so I could not solve it.

I'd appreciate your help.

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You need not find the inverse. The computation simply requires application of the chain rule.

Observe that $$g(x)=f^{-1}(x)$$ $$\Rightarrow f\{g(x)\}=x$$

Now, differentiating both sides with respect to $x$, using chain rule, we get

$$\frac{d}{dx}f\{g(x)\}=\frac{d}{dx}x$$ $$\Rightarrow \frac{d}{dg(x)}f\{g(x)\}\cdot \frac{dg(x)}{dx}=1$$ $$\Rightarrow f'\{g(x)\}\cdot g'(x)=1$$ $$\Rightarrow g'(x)=\frac{1}{f'\{g(x)\}}$$

Hence $$g'(1)=\frac{1}{f'\{g(1)\}}$$

Hope you can finish this on your own.

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Since $g\bigl(f(x)\bigr)=x$, the chain rule implies that$$g'\bigl(f(x)\bigr)f'(x)=1.$$In particular, put $x=0$ and you will get:$$g'(1)f'(0)=1.$$

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Use the chain rule to differentiate both sides of either $$f(g(x))=x$$or $$g(f(x))=x$$whichever seems easiest to you.

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  • $\begingroup$ please tell me about the chain rule $\endgroup$ – Usama Enany Dec 22 '17 at 17:20
  • $\begingroup$ If you don't know about the chain rule, and the problem tells you to use the chain rule, then your first order of business, before coming here to post the question, should have been to look for resources on the chain rule, either in whatever book you're using, or via google. This problem is meant for people who have already mastered the chain rule on simpler examples (like differentiating $e^{3x^2+2x}$ or $\sqrt{\ln x}$) and are ready to take it to a slightly more abstract level. $\endgroup$ – Arthur Dec 22 '17 at 17:24
  • $\begingroup$ I know about the chain rule but I just couldn't see how it could be applied here. Anyways I know recognize it. $\endgroup$ – Usama Enany Dec 22 '17 at 17:40
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A nice little trick:

Notice the following

$$\frac{d}{dx} f(f^{-1}(x)) = \frac{d}{dx}(f^{-1}(x)) \cdot f'(f^{-1} (x))$$

where the equality comes from the chain rule. On the left hand side, however, we know that

$$f(f^{-1}(x)) = x$$

because they are inverse functions. Therefore,

$$\frac{d}{dx} f(f^{-1}(x)) = \frac{d}{dx} x = 1$$

So putting this into the first equation, we have that

$$1 = \frac{d}{dx}(f^{-1}(x)) \cdot f'(f^{-1} (x))$$

$$\implies \frac{d}{dx}(f^{-1}(x)) = \frac{1}{f'(f^{-1} (x))}$$

Since $g(x) = f^{-1}(x)$ then we have

$$g'(x) = \frac{1}{f'(g(x))}$$

$$\implies g'(1) = \frac{1}{f'(g(1))}$$

If we do a bit of a guess, we see that $f(0) = 1$. I'll leave it from here for you to do.

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Use the formula $(f^{-1})^{'}(x) = \frac 1 {f^{'}(f^{-1}(x))}$ then use the fact that $f^{-1}(1)=0$

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This function hasn't an inverse function that you can describe using elementary functions. If you have a given value, $ y_0 $ and you want to know for which $ x_0 $ is true the equation $ y_0 = f(x_0) $ you may use the Newton's method: Newton's method in Wikipedia.

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It is true the function $f$ is not one-to-one for all $x$. Instead one must restrict its domain in order for the inverse to exist at the point of interest of $x = 1$.

As the function $f$ has a single global minimum at the point $$x^* = -\frac{1}{2} - \text{W}_0 \left (\frac{1}{2 \sqrt{e}} \right ) = -0.738\,835\ldots,$$ where $\text{W}_0(x)$ is the principal branch of the Lambert W function, by restricting the domain of the function $f$ to $x \geqslant x^*$ an inverse $g (x) = f^{-1}(x)$ containing the point $x = 1$ exists though an explicit expression for it cannot be found.

It may however still be possible to find individual values for $f^{-1} (x)$ at particular values for $x$. In our case we note that $f^{-1} (1) = 0$. This allows for the value of the derivative at this point to be found using $$(f^{-1})'(1) = \frac{1}{f'(f^{-1} (1))} = \frac{1}{f'(0)},$$ since we know that $$(f^{-1})' (x) = \frac{1}{f'(f^{-1} (x))},$$ and the value $f'(0)$ can be readily found.

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