How do you differentiate #f(x)=x^3*sqrt(x-2)-sinxcosx# using the product rule?

Question

Charles Anctil · Accepted Answer

#(7x^3 - 12x^2)/(2 sqrt(x-2)) + sin^2x - cos^2x#

#f(x)# is the sum of two functions, each of which is the product of two other functions (hence the requirement for the product rule).

Starting with the first of the two products,

denoting #g(x) = x^3# and #h(x) = sqrt(x - 2)#

the product rule states

#f'(x) = (g(x)h(x))' = g'(x)h(x) + g(x)h'(x)#

#g'(x) = 3x^2# (simple polynomial differentiation)

#h'(x)# is trickier because it is a compound function (function of a function) which requires the chain rule. It is necessary to specify more functions.

Detour for chain rule applied to the second function:

Denoting

#j(x) = sqrt(x)# and #k(x) = x - 2#

this implies #h(x) = j(k(x))#

under the chain rule,

#h'(x) = j'(k(x))k'(x)#

noting #sqrt(x) = (x)^(1/2)#

#j'(x) = 1/2 (x^(-1/2))#

so that

#j'(k(x)) = 1/2 ((x - 2)^(-1/2))#

Also

#k'(x) = 1#

so

#h'(x) = j'(k(x))k'(x) = 1/2 ((x - 2)^(-1/2))(1)#

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

or, reverting to square root notation,

#h'(x)= 1/(2 sqrt(x-2))#

Returning to the application of the product rule

#f'(x) = (g(x)h(x))' = g'(x)h(x) + g(x)h'(x)#

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

#= (6x^2(x - 2) + x^3)/(2 sqrt(x-2))#

#= (6x^3 - 12x^2 + x^3)/(2 sqrt(x-2))#

#= (7x^3 - 12x^2)/(2 sqrt(x-2))#

On to the second of the two products.

Denoting #m(x) = sin(x)# and #n(x) = cos(x)#

as already noted, under the product rule,

#(m(x)n(x))' = m'(x)n(x) + m(x)n'(x)#

#m'(x) = cos(x)# and #n'(x) = - sin(x)#

so

#(m(x)n(x))' = m'(x)n(x) + m(x)n'(x) = cos(x)cos(x) + sin(x)(- sin(x))#

that is

#(m(x)n(x))' = cos^2(x) - sin^2(x)#

So, the overall derivative (under the sum rule) is

#f'(x) = (7x^3 - 12x^2)/(2 sqrt(x-2)) - (cos^2(x) - sin^2(x))#

#f'(x) = (7x^3 - 12x^2)/(2 sqrt(x-2)) + sin^2(x) - cos^2(x)#

Evelyn Agard · Answer

undefined To differentiate $ f(x) = x^3\sqrt{x-2} - \sin(x)\cos(x) $ using the product rule, first, identify the functions being multiplied together. Then, apply the product rule which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Let $ u = x^3 $ and $ v = \sqrt{x-2} $, and $ w = \sin(x) $ and $ z = \cos(x) $.

Then, differentiate each function with respect to $ x $:

$ u' = 3x^2 $ (derivative of $ x^3 $)

$ v' = \frac{1}{2\sqrt{x-2}} $ (derivative of $ \sqrt{x-2} $)

$ w' = \cos(x) $ (derivative of $ \sin(x) $)

$ z' = -\sin(x) $ (derivative of $ \cos(x) $)

Now, apply the product rule:

$ f'(x) = u'v + uv' - w'z - wz' $

$ f'(x) = (3x^2)(\sqrt{x-2}) + (x^3)\left(\frac{1}{2\sqrt{x-2}}\right) - (\sin(x))(-\sin(x)) - (\cos(x))(\cos(x)) $

Simplify the expression to obtain the derivative of $ f(x) $.