How do you find the derivative of #4sqrt(x-3) + x^5/(2x)#?

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Answer 1 · 2015-05-20T16:38:50

The first step here would be to write the equation in a way that is easier to understand:

#f(x) =4sqrt(x-3) + x^5/(2x) = 4(x-3)^(1/2) + x^4/2#

now we solve for #f'(x)#

we can use the chain rule in the first section. The chain rule states that:

#(dy)/(du)*(du)/(dx)=(dy)/(dx)#

Naming #u=(x-3)# and then using the chain rule, we get:

#color(red)(d/(dx) 4(x-3)^(1/2) = 2(x-3)^(-1/2)(1) = 2(x-3)^(-1/2))#

and the second section would be:

#color(blue)(d/dx x^4/2 = 2x^3)#

now we also know that

#f'(x) = d/(dx) 4(x-3)^(1/2) + d/dx x^4/2#

so we can now just drop in our two results:

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