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

Question

Scarlett Allison · Accepted Answer

# f'(x) = (5x^2 - 8x)/(2sqrt(x - 2)) #

using the 'product rule' and the 'chain rule ' : rewrite f(x) = # x^2sqrt(x - 2) = x^2.(x - 2 )^(1/2) # f'(x) = #x^2 d/dx(x - 2 )^(1/2) + (x - 2 )^(1/2) d/dx (x^2) # #= x^2(1/2(x - 2 )^(-1/2) d/dx(x - 2 ) )+ (x - 2 )^(1/2) .(2x)# #= x^2(1/2 (x - 2 )^(-1/2) . 1 ) + 2x(x - 2 )^(1/2)# # = 1/2 x^2 (x - 2 )^(-1/2) + 2x (x - 2 )^(1/2) # [ common factor of #(x - 2 )^(-1/2)# ] #= (x - 2 )^(-1/2) [1/2 x^2 + 2x(x - 2 ) ]# # = (x - 2 )^(-1/2) [ 1/2 x^2 + 2x^2 - 4x ] <# = (x - 2 )^(-1/2) [ 1/2 (x^2 + 4x^2 - 8x )] # #rArr f'(x) = (x - 2 )^(-1/2) 1/2 ( 5x^2 - 8x ) = (5x^2 - 8x )/(2sqrt(x - 2 ))#

Emilia Aguilar · Answer

undefined To differentiate the function $ f(x) = x^2 \cdot \sqrt{x-2} $ using the product rule, we first identify two functions: $ u = x^2 $ and $ v = \sqrt{x-2} $. Then, we differentiate each function with respect to $ x $ to find $ u' $ and $ v' $. Applying the product rule formula $ (uv)' = u'v + uv' $, we calculate:

$$ u' = 2x $$
$$ v' = \frac{1}{2\sqrt{x-2}} $$

Now, we substitute these values into the product rule formula:

$$ f'(x) = (x^2)' \cdot \sqrt{x-2} + x^2 \cdot (\sqrt{x-2})' $$

$$ f'(x) = (2x) \cdot \sqrt{x-2} + x^2 \cdot \frac{1}{2\sqrt{x-2}} $$

$$ f'(x) = 2x \cdot \sqrt{x-2} + \frac{x^2}{2\sqrt{x-2}} $$

$$ f'(x) = 2x\sqrt{x-2} + \frac{x^2}{2\sqrt{x-2}} $$