Question

Differentiate:
f(x)=sqrt(x) lnx

Answer 1

Use product rule.

Answer 2

Use chain [d/dx f(g) = f'(g)g'] and product rule [d/dx fg = f'g + fg']. Let:

f = √x = x^(½)
f' = ½(x)^(½ - 1) [Chain Rule]
= ½(x)^(½ - 2/2)
= ½(x)^(-½)
= 1/(2√x)

g = ln(x)
g' = 1/x [Chain Rule again; Note that d/dx ln(x) = 1/x]

So:

f'(x) = ln(x)/(2√x) + √(x)/x
= ln(x)/(2√x) + x^(½)/x^(1)
= ln(x)/(2√x) + 1/x^(1 - ½)
= ln(x)/(2√x) + 1/x^(½)
= ln(x)/(2√x) + 1/√x

Answer 3

That's not the chain rule, that's the power rule.

Answer 4

Ahhhh THank You.

Answer 5

so im right

Answer 6

Anything can be written as a chain rule. \[\frac{ d }{ dx } x ^{2}=\frac{ d }{ dx }(x)^{2}=2(x)\frac{ d }{ dx }(x)=2x \times1=2x\]

Answer 7

Well, not everything, but the point is there.

Answer 8

\[\frac{d(\sqrt{x}\ln(x))}{dx}=\frac{1}{2}\frac{\ln(x)}{\sqrt x}-\frac{1}{\sqrt x}\]

Answer 9

is this the function you want to differentiate...

Answer 10

i just noticed pooja195 his answer is correct...

Answer 11

Yes, thanks to everyone for clarifying this one. I think I got it.