Question

Find the derivative using the derivative rules?

Answer 1

See explanation.

`(1-72x^2)/sqrt(1-36x^2)`

Explanation:

There appears to be an error on the problem, specifically the line right after the exponent. It doesn't appear to be a 1, and if it is a symbol, I do not recognize it. This explanation will continue under the presumption that the extra line present in the problem is an error. It will be presumed that the function is `y=xsqrt(1-36x^2)`

We must use both the Product Rule and the Chain Rule for this.

Product Rule: `y=f(x)g(x), dy/dx = f'(x)g(x) + f(x)g'(x)`

Chain Rule: Given `h(k(x)), (dh)/(dx) = (dk)/dx * (dh)/(dk)`

We have `f(x) = x, f'(x) = 1, g(x) = sqrt(1-36x^2)`

g(x) will break down into our h and k...

`k(x) = 1-36x^2, (dk)/dx = -72x, h(k) = sqrtk = k^(1/2), (dh)/(dk) = 1/(2k^(1/2)) = 1/(2sqrtk) = 1/(2sqrt(1-36x^2))`

Thus:

`dy/dx = (1)sqrt(1-36x^2) + (x) (dk)/dx (dh)/(dk) = sqrt(1-36x^2) +(x)(-72x)(1/(2sqrt(1-36x^2))) = sqrt(1-36x^2)- (36x^2)/sqrt(1-36x^2)`

We can simplify this...

`= (1-36x^2)/sqrt(1-36x^2) - (36x^2)/sqrt(1-36x^2) = (1-72x^2)/sqrt(1-36x^2)`

Note that, provided we're working with real numbers, the domain is rather small. We must have `1-36x^2>=0` , which we can quickly determine means we must have `-1/6 <= x <= 1/6`

Answer 2

Use the product rule and chain rule.
`f'(x)=(1-72x^2)/(sqrt(1-36x^2))`

Explanation:

Chain rule: `F(x)=f'(g(x))*g'(x)`
Product Rule: `F(x)=f'(x)*g(x)+g'(x)*f(x)`
Reduce the square root to a fractional exponent.
`sqrtx=x^(1/2)`
Where the `f(x)=x`; `g(x)=sqrt(1-36x^2)` or `(1-36x^2)^(1/2)`
Have fun!