How do you differentiate #g(y) =xsqrtx # using the product rule?

Question

How do you differentiate #g(y) =xsqrtx   # using the product rule?

Ezra Adams · Accepted Answer

For #g(x)=xsqrtx#, we get #g'(x)=(3sqrtx)/2#

For #g(x)=xsqrtx = x^(3/2)#, but the question specifies the product rule, so we think of it as:

#g(x)=FS#.

The product rule gives #g'(x) = F'G+FG'#.

Here, we have #F=x#, so F'=1#

and #G=sqrtx=x^(1/2)#, so #G' = 1/2 x^(-1/2) = 1/(2sqrtx)#.

When we use the product rule, we obtain

#g'(x) = (1)(sqrtx)+(x)(1/(2sqrtx))#

# = sqrtx + x/2sqrtx =sqrtx + sqrtx/2 = 3/2 sqrtx#

Cameron Alexander · Answer

undefined To differentiate $ g(y) = x \sqrt{x} $ using the product rule:

1. Identify the functions: $ f(x) = x $ and $ h(x) = \sqrt{x} $.
2. Compute the derivatives of each function: $ f'(x) = 1 $ and $ h'(x) = \frac{1}{2\sqrt{x}} $.
3. Apply the product rule: $ g'(x) = f(x)h'(x) + f'(x)h(x) $.
4. Substitute the derivatives and functions: $ g'(x) = x \cdot \frac{1}{2\sqrt{x}} + 1 \cdot \sqrt{x} $.
5. Simplify the expression: $ g'(x) = \frac{1}{2\sqrt{x}} \cdot x + \sqrt{x} $.
6. Further simplify if needed: $ g'(x) = \frac{x}{2\sqrt{x}} + \sqrt{x} $.
7. Finally, simplify the expression: $ g'(x) = \frac{x}{2\sqrt{x}} + \sqrt{x} = \frac{1}{2} \sqrt{x} + \sqrt{x} = \frac{3}{2} \sqrt{x} $.