I have a math problem where I am required to find the derivative of a function with the limitations of not being allowed to use the Product or Quotient Rule of Differentiation.
The problem looks like this:
$$h(x) = \frac{4-x^6}{3x^{-2}}$$
I have tried a variety of routes but always end up with results that seem to require the use of the Product or Quotient Rule.
For example, my latest try looks like this:
$$h(x) = \frac{4-x^6}{3x^{-2}}$$
$$h(x) = \frac{4}{3x^{-2}} - \frac{x^6}{3x^{-2}}$$
$$h(x) = \frac{4x^2}{3} - \frac{x^8}{3}$$
(From this step, I figured I could just use the Difference Rule of Differentiation, like this:)
$$h'(x) = \frac{d}{dx}\left(\frac{4x^2}{3}\right) - \frac{d}{dx}\left(\frac{x^8}{3}\right)$$
But wouldn't this actually end up using the Product -or- Quotient Rule? Like this:
$$h'(x) = \frac{d}{dx}\left(\frac{4}{3}(x^2)\right) - \frac{d}{dx}\left(\frac{1}{3}(x^8)\right)$$
Is there another route I can take with this type of problem that would avoid using the Product or Quotient Rule of Differentiation?
$\endgroup$ 32 Answers
$\begingroup$Most likely you are allowed to use the Constant Multiple Rule, i.e. $(cf(x))'=c(f(x))'$, where $c$ is a constant.
$\endgroup$ 1 $\begingroup$Let $$h(x)=f(x)\cdot g(x)$$
So, $$\ln h(x)=\ln f(x)+\ln g(x)$$
Differentiating wrt $x$ using Chain Rule,
$$\frac{h'(x)}{h(x)}=\frac{f'(x)}{f(x)}+\frac{g'(x)}{g(x)}$$
$$\implies h'(x)=\frac{h(x)}{f(x)}\cdot f'(x)+\frac{h(x)}{g(x)}\cdot f'(x)=g(x)\cdot f'(x)+f(x)\cdot g'(x)$$
If $f(x)=c,$ (constant)
$f'(x)=0$ and $h(x)=c\cdot g(x)\implies h'(x)=c\cdot g'(x)$
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