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Consider this:

Find the tangent to the function $f$ at point $(2,f(2))$ where $f(2)=0$ and the function $f:(1,\infty)\to\mathbb{R}$ using the implicit equation:

$$xe^{xf(x)} = (x-2)^3 + e^{f(x)}+1.$$

Now I've found tried to find the $g'(x)$ and $h'(x)$.

$$g(x) = xe^{xf(x)} \implies g'(x) = e^{xf(x)} + xe^{xf(x)'} * f(x)';$$ $$f(x) = (x-2)^3 + e^{f(x)} +1 \implies h'(x) = 3(x-2)^2 + e^{f(x)'}. $$

Now I am not sure if I derived correctly, and if not please let me know. But the other important thing is I'm not sure what to do next? Am I supposed to input $2$ into $f(x)$ to "simplify". A tip would be great.

Thanks!

amWhy
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codeisfun
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1 Answers1

2

You're wrong in a couple of points:
$\frac{d}{dx}xe^{xf(x)}=e^{xf(x)}+ x\bigg(e^{xf(x)}\cdot \big(xf'(x)+f(x)\big)\bigg)=$
$\frac{d}{dx}\bigg((x-2)^3+e^{f(x)}+1\bigg)=3(x-2)^2+e^{f(x)}\cdot f'(x)$
So finally we can compute the derivative: (Just factorize $f'(x)$ out of both sides of the equality) $$f'(x)=\frac{e^{xf(x)}\big(xf(x)+x+1\big)-3(x-2)^2}{(1-x^2)e^{f(x)}}$$
Now because $f(2)=0$, clearly $f'(2)=\frac{3}{-3}=-1.$ So the slope of the tangent line is $-1$ and it also passes the point $(2,0)$ so the equation of the tangent line is: $y=2-x$ and the problem is solved.

Aryan
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