# Using the power rule instead of the reciprocal rule?

I'm stuck with this problem: I got to find the derivative of $$\frac{1}{x^2+80}$$ and I do know how to solve it by using the reciprocal rule, but whenever I try to solve it using the power rule I get a discrepant result:

$$\frac{d}{dx}$$ $$(x^2+80)^-1$$ = $$-(t^2+80)^-2$$ = $$-\frac{1}{(t^2+80)^2}$$

When the result really should be equal to:

$$-\frac{2x}{(t^2+80)^2}$$

What am I doing wrong?

• Always remember the chain rule! The inside function is $x^2 + 80$ whose derivative is $2x$. Also, don't confuse $t$ and $x$ – Andrew Li Mar 1 at 2:09

## 2 Answers

I've never heard of a 'reciprocal rule,' usually this is just the power rule for negative exponents. The thing here you've forgotten is the chain rule - you took the derivative of the outside correctly, but when you have a composition of functions, you have to multiply by the derivative of the inside.

So you need to calculate $$\frac{d}{dx} (x^2 + 80) = 2x$$ and multiply that in, hence the factor.

Also you need to decide if you want $$t$$ or $$x$$ as your variable and stick to it.

• And why is $\frac{1}{x^2+80)}$ a composition of functions? Can't it just be $f(x)=\frac{1}{x^2+80}$? – Christus Moura Mar 2 at 4:04
• This is the composition of functions $f(x) = x^2 + 20$ and $g(y) = 1/y$. The deal is that you use the chain rule whenever you can write your function as a composition of elementary functions such as these. – Alfred Yerger Mar 2 at 4:59
• If you wanted to write it that way though, you could use the quotient rule of course. – Alfred Yerger Mar 2 at 5:14

You are forgetting the chain rule:

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

Now let $$f(y) = y^{-1}$$ and $$g(x)=x^2+80$$ thus $$f(g(x)) = (x^2+80)^{-1}$$ which is your desired function. I'll let you take it from here.

It is also important to note that the reciprocal rule is equivalent to the chain rule since the reciprocal rule is:

$$\frac{d}{dx}(\frac{1}{f(x)}) = \frac{-f'(x)}{(f(x))^2}$$

and the chain rule easily derives this:

$$\frac{d}{dx}(\frac{1}{f(x)}) = \frac{d}{dx}(f(x))^{-1}=-f(x)^{-2}f'(x)=\frac{-f'(x)}{(f(x))^2}$$