Question

can someone help me find the derivative of a square root step by step please.

Answer 1

Sqrt(2-5*x^2)

Answer 2

i did this: 2-5x^2) ^1/2

Answer 3

then 1/2(2-5x^2)

Answer 4

@zepdrix

Answer 5

Woops you didn't apply the power rule correctly.\[\large (2-5x^2)^{1/2}\]Derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{1/2-1} \qquad=\qquad \frac{1}{2}(2-5x^2)^{-1/2}\]But we also have to apply the chain rule after that, multiplying by the derivative of the inside.

Answer 6

uh-oh got me there... may i ask how?

Answer 7

i think i kind of do but im not sure how to proceed it.

Answer 8

\[\large \color{royalblue}{\left[(2-5x^2)^{1/2}\right]'} \qquad = \qquad \frac{1}{2}(2-5x^2)^{-1/2}\color{royalblue}{(2-5x^2)'}\]We took the derivative of the `outer` function, apply the power rule.
Now we have to multiply by the derivative of the inside. The prime is to show that we still need to take the derivative of that part.
Taking that derivative gives us,\[\large \frac{1}{2}(2-5x^2)^{-1/2}\color{orangered}{(-10x)}\]

Answer 9

Confused? :O Too much stuff going on?

Answer 10

im a little confused with how you got to 10x

Answer 11

It's the derivative of that blue thing.
Take the derivative of each term inside the brackets, separately.

Answer 12

isn't it this: f(x)'*(g(x)+f(x)*g(x)' ?

Answer 13

i see but what happens to the half?

Answer 14

That would be the product rule:\[\large \left[f(x) \cdot g(x)\right]' \qquad=\qquad f'(x)g(x)+f(x)g'(x)\]

We're dealing with a composition of functions, a function within a function.
Which tells us to apply the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]

Answer 15

The chain rule tells us...
After you've taken care of the outer function...
(In our case our outer function is "something" to the 1/2 power)
make a copy of the inside, take it's derivative, and multiply by it.

Answer 16

I hate using function notation like this, but maybe it'll help you to understand the chain rule.
I find it more confusing, but to each his own.

So we have this,\[\large f(x)=x^{1/2}\]\[\large g(x)=2-5x^2\]

\[\large f(g(x))=(2-5x^2)^{1/2}\]

Taking the derivative using the chain rule:\[\large \left[f(g(x))\right]' \qquad=\qquad f'(g(x))g'(x)\]

\[\large =\left[(2-5x^2)^{1/2}\right]'\left[2-5x^2\right]'\]

Answer 17

If we have a function of a potato...
The chain rule tells us that after we take the derivative of the function, we make a copy of the potato and take it's derivative as well.