avendanl 2 years ago Differentiate: f(x)=sqrt(x) lnx

1. wio

Use product rule.

2. pooja195

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

3. wio

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

4. avendanl

Ahhhh THank You.

5. pooja195

so im right

6. calmat01

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

7. calmat01

Well, not everything, but the point is there.

8. mathsmind

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

9. mathsmind

is this the function you want to differentiate...

10. mathsmind

i just noticed pooja195 his answer is correct...

11. avendanl

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