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How would I begin solving a problem that is asking me to find the derivative of $F$ when $$F(x)=\left(\frac{1}{3x-f(x)}\right)^4$$ and $f$ is differentiable.

Not asking for the answer here obviously, just the steps needed to get off the ground.

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$$F(x)=\left(\frac{1}{3x-f(x)}\right)^4$$

Because if $F=g(x)^n$ then $F'=ng(x)^{n-1}g'(x)$

$$F'(x)=4(\frac{1}{3x-f(x)})^3(\frac{1}{3x-f(x)})'$$

Here the derivative of $ \frac 1 {g(x)}$ is $ \frac {-g(x)'} {g(x)^2}$

$$F'(x)=4(\frac{1}{3x-f(x)})^3(\frac{-3+f'(x)}{(3x-f(x))^2})$$

Therefore,

$$F'(x)=4\frac{(-3+f'(x))}{(3x-f(x))^5}$$

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$$F(x)=\left(\frac{1}{3x-f(x)}\right)^4=\frac{1^4}{(3x-f(x))^4}=(3x-f(x))^{-4}$$ Then, $$\frac{d}{dx}F(x)=-4\cdot(3x-f(x))^{-5}\cdot(3-f'(x))$$

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$F'(x) = -4(13-f'(x))(13x-f(x))^{-5}$ by the Chain Rule

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  • $\begingroup$ Sorry, I mistyped my question. You can look at the revised version above. $\endgroup$ – Ulysses Li Oct 13 '17 at 21:58
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For problems in which appear products, quotients and powers, logarithmic differentiation make life much easier.

To make the problem more complex, let me consider $$F(x)=\frac{\left(a(x)+ b(x)\right)^m}{\left(c(x)+ d(x)\right)^n}$$ Take logarithms $$\log(F(x)=m\log\left(a(x)+ b(x)\right)-n\log\left(c(x)+ d(x)\right)$$ Differentiate $$\frac{F'(x)}{F(x)}=m\frac {a'(x)+ b'(x)}{a(x)+ b(x)}-n\frac {c'(x)+ d'(x)}{c(x)+ d(x)}$$ Then, use $$F'(x)=F(x)\times \left(\frac{F'(x)}{F(x)} \right)$$ and simplify.

If you thinks about it, it is the cheapest way to use if you have to code it.

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