How do you differentiate #sqrt(2x+1)(x^2+1)#?

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Answer 1 · 2017-08-31T20:20:07

Break it down into pieces, and derive each using rules you already know...

Start by recognizing that what you have is the product of 2 functions.

The rule for finding the derivative of the product of 2 functions is:

#d/dx(f(x)g(x)) = d/dxf(x) * g(x) + f(x) * d/dx(g(x))# (we'll label this as eq. 1)

here, #f(x) = sqrt(2x + 1)#, and #g(x) = x^2 + 1#

You can use the chain rule for deriving f(x). Start by re-writing it as

#f(x) = (2x + 1)^(1/2)#

#(df)/dx = (1/2)(2x + 1)^(-1/2) * 2 = 1/sqrt(2x + 1)#

The derivative of #g(x) = x^2 + 1# is simple - it's just #2x#

Plug all this back into eq. 1:

#d/dx(f(x)g(x)) = (x^2 + 1)/sqrt(2x + 1) + (sqrt(2x + 1) * 2x)#

...which is your answer, but your instructor will probably want you to simplify, because they're like that.

Multiply the right second term by #sqrt(2x + 1)/sqrt(2x + 1)# to get a common denominator:

#= (x^2 + 1)/sqrt(2x + 1) + ((2x + 1) * 2x)/sqrt(2x + 1)#

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

# = (5x^2 + 2x + 1)/sqrt(2x + 1)#

...and turn that puppy in.

GOOD LUCK