Negative exponents when multiplying polynomials

Question

$$(4-t)(1+t^2)^{-1}$$ I am supposed to find the derivative of this but I am not sure if it means $$\frac{1}{(4-t)(1+t^2)^{-1}}\quad\text{or}\quad\frac{(4-t)}{(1+t^2)}$$ I have tried to look online for help but couldn't find an example that looked like this. I am assuming that the second one is the correct one but I just wanted to make sure.

Answer 1

It is the last of the two: the scope of the negative exponent applies only to the factor $(1 + t^2)$.

With that, your function, we'll call $h(x)$ is written as follows:
$$h(x) = (4-t)(1+t^2)^{-1} = \frac{(4-t)}{(1+t^2)}$$

• If your function were expressed as $\displaystyle\left[(4-t)(1+t^2)\right]^{-1}$, then the scope of the negative exponent would be over both factors, and so $\displaystyle \left[(4-t)(1+t^2)\right]^{-1}$ would be equivalent to $\dfrac{1}{(4-t)(1+t^2)}$

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You can apply the quotient rule for finding the derivative of $$h(x) = \frac{(4-t)}{(1+t^2)}$$

Let $f(x) = 4 - t$, and let $g(x) = (1 + t^2).\;\;$ Then $$h'(x) = \frac{f(x)g'(x) - f'(x)g(x)}{g^2(x)}$$

I trust you can take it from here?

Answer 2

Implicit multiplication is not an expression of containment! Therefore, the $^{-1}$ does not apply to both factors, but only to the second.