How do you find the derivative of #3*(sqrtx) - (sqrtx^3)#?

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Answer 1 · 2016-02-10T02:53:56

#(3(1-x))/(2sqrtx)#

Call the function #f(x)#. Now, write the function using fractional exponents instead of with radicands:

#f(x)=3x^(1/2)-(x^(1/2))^3#

Multiply the exponents in the second term to get:

#f(x)=3x^(1/2)-x^(3/2)#

Now, each term can be differentiated through the power rule, which states that

#d/dx(x^n)=nx^(n-1)#

Recall that constants being multiplied simply stay being multiplied. Applying the power rule to each term gives a derivative of

#f'(x)=3(1/2)x^(1/2-1)-3/2x^(3/2-1)#

Simplify.

#f'(x)=3/2x^(-1/2)-3/2x^(1/2)#

While this is a fine final answer, it's often helpful to put functions like this in fractional form so the derivative can be easily set equal to #0#:

#f'(x)=(3/2x^(-1/2)-3/2x^(1/2))/1(x^(1/2)/x^(1/2))#

#f'(x)=(3-3x)/(2x^(1/2))#

#f'(x)=(3(1-x))/(2sqrtx)#