Find the derivative using the derivative rules?

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2 Answers
Dec 7, 2017

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#

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!