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What is the derivative of #sqrt(4x² + 1)#?

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Ben Share
Mar 20, 2018

Answer:

#d/(dx)=(4x)/(sqrt(4x^2+1))#

Explanation:

The chain rule has you treat functions as follows:

#f(x)=g(x)^n#

#f'(x)=(n*g(x)^(n-1))*g'(x)#

You treat anything inside of the exponent as a second function, and you first take the derivative of the exponent, and then multiply by the derivative of the inner function.

For this problem:

#g(x)=4x^2+1#

#f(x)=sqrt(g(x))=(g(x))^(1/2)#

Now, lets take our derivatives:

#g'(x)=8x#

#f'(x)=1/2(g(x))^(-1/2)*g'(x)#

#f'(x)=(1/2)1/(g(x))^(1/2)*g'(x)#

#f'(x)=(1/2)1/sqrt(4x^2+1)*(8x)#

#color(red)(f'(x)=(4x)/sqrt(4x^2+1)#

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