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

Question

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

Scarlett Allison · Accepted Answer

The answer is #=5/2xsqrtx#

The rule for products is

#(uv)'=u'v+uv'#

Here, we make use of

#(x^n)'=nx^(n-1)#

#(sqrtx)'=1/(2sqrtx)#

So,

#u=x^2#, #=>#, #u'=2x#

#v=sqrtx#, #=>#, #v'=1/(2sqrtx)#

Consequently,

#f'(x)=2x*sqrtx+x^2*1/(2sqrtx)#

#=(4x^2+x^2)/(2sqrtx)#

#=(5x^2)/(2sqrtx)=5/2xsqrtx#

Emily Ammerman · Answer

undefined To differentiate $ f(x) = x^2 \sqrt{x} $ using the product rule:

1. Identify the two functions being multiplied: $ f(x) = u(x) \cdot v(x) $, where $ u(x) = x^2 $ and $ v(x) = \sqrt{x} $.
2. Apply the product rule: $ f'(x) = u'(x)v(x) + u(x)v'(x) $.
3. Find the derivatives of each function:
   - $ u'(x) = 2x $ (derivative of $ x^2 $).
   - $ v'(x) = \frac{1}{2\sqrt{x}} $ (derivative of $ \sqrt{x} $).
4. Substitute the derivatives and original functions into the product rule formula:
   $ f'(x) = (2x) \cdot \sqrt{x} + (x^2) \cdot \frac{1}{2\sqrt{x}} $.
5. Simplify the expression:
   $ f'(x) = 2x\sqrt{x} + \frac{x^2}{2\sqrt{x}} $.
6. If needed, further simplify the expression by rationalizing the denominator:
   $ f'(x) = 2x\sqrt{x} + \frac{x^2\sqrt{x}}{2x} $.
   $ f'(x) = 2x\sqrt{x} + \frac{x\sqrt{x}}{2} $.