Question

Differentiate `\frac{r}{\sqrt{r^2+5}}` ?

I'm unsure how they went from the first step to the second step? How did the `(r^2+5)^(1/2)` on the denominator from the first step turn into an exponent of 1 on the second step? Also, how do you get the -r^2 in the numerator in the second step?

Answer 1

`5/(sqrt((r^2+5)^3)`

Explanation:

`"differentiate using the "color(blue)"product rule"`

`"given "y=g(h(x))" then"`

`dy/dx=g(x)h'(x)+h(x)g'(x)larrcolor(blue)"product rule"`

`"here "y=r/(sqrt((r^2+5)^(1/2)))=r(r^2+5)^(-1/2)`

`g(r)=r rArrg'(r)=1`

`h(r)=(r^2+5)^(-1/2)`

`h'(r)=-1/2(r^2+5)^(-3/2)xxd/(dr)(r^2+5)`

`color(white)(h'(x))=-1/2(r^2+5)^(-3/2)xx2r=-r(r^2+5)^(-3/2)`

`(dy)/(dr)=-r^2(r^2+5)^(-3/2)+(r^2+5)^(-1/2)`

`color(white)((dy)/(dr))=(r^2+5)^(-3/2)[cancel(-r^2)cancel(+r^2)+5]`

`color(white)((dy)/(dr))=5/(r^2+5)^(3/2)`

Answer 2

By expanding by `sqrt(r^2+5)/sqrt(r^2+5)`

Explanation:

Your first question can be answered by a simple algebra rule. We know that :

`(a^m)^n = a^(m*n)`

so therefore `((r^2+5)^(1/2))^2 = (r^2+5)^(2*1/2) = (r^2+5)`

In the next step, we expanded the fraction by `sqrt(r^2+5)/sqrt(r^2+5)`.

Therefore in the denominator we got `(r^2+5)^(3/2)` and the second part of the numerator became :

`-r^2*(r^2+5)^(-1/2)*(r^2+5)^(1/2) = -r^2`.