How do you find the derivative of #f(x) = x^2 (x-5)^3#?

Question

Charles Anctil · Accepted Answer

#(d f(x))/(d x)=2x(x-5)^3+3x^2(x-5)^2# #(d f(x))/(d x)=?# #y=a*b" ; "y'=a'*b+b^'*a# #(d f(x))/(d x)=2x*(x-5)^3+3(x-5)^2*1*x^2# #(d f(x))/(d x)=2x(x-5)^3+3x^2(x-5)^2#

Isaiah Allen · Answer

Step 1:  Determine the derivative of #(x - 5)^3# using the chain rule Let #y  =u^3# and #u = x - 5#.  The derivative of #y# is #y' = 3u^2# and the derivative of #u# is #1#.  Hence, the derivative of #y = (x - 5)^3# is #y' = 3(x - 5)^2 xx 1 =color(red)( 3(x - 5)^2)# Step 2:  Determine the derivative of #f(x)# using the product rule Let #f(x) = g(x) xx h(x)#.  Then our derivative is given by #f'(x) = g'(x) xx h(x) + g(x) xx h'(x)#.   The derivative of #g(x)# is #2x#.  The derivative of #h(x)#, as mentioned in step #1# is #3(x - 5)^2#. Hence, #f'(x) = 2x xx(x - 5)^3 + x^2 xx 3(x - 5)^2 =2x(x - 5)^3 + 3x^2(x - 5)^2 = (x - 5)^2(2x(x - 5) + 3x^2) = (x - 5)^2(2x^2 - 10x + 3x^2) = (x - 5)^2(5x^2 - 10x) =color(blue)( 5x(x - 5)^2(x - 2))# I hope this is useful!

Evelyn Agard · Answer

undefined To find the derivative of $ f(x) = x^2 (x-5)^3 $, you can use the product rule:

$$ \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) $$

Where $ f(x) = x^2 $ and $ g(x) = (x-5)^3 $.

Now, apply the product rule:

$$ f'(x) = \frac{d}{dx}(x^2) = 2x $$
$$ g'(x) = \frac{d}{dx}[(x-5)^3] = 3(x-5)^2 $$

Therefore, the derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = 2x \cdot (x-5)^3 + x^2 \cdot 3(x-5)^2 $$