Here's the question you clicked on:
avendanl
Differentiate: f(x)=sqrt(x) lnx
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
That's not the chain rule, that's the power rule.
Anything can be written as a chain rule. \[\frac{ d }{ dx } x ^{2}=\frac{ d }{ dx }(x)^{2}=2(x)\frac{ d }{ dx }(x)=2x \times1=2x\]
Well, not everything, but the point is there.
\[\frac{d(\sqrt{x}\ln(x))}{dx}=\frac{1}{2}\frac{\ln(x)}{\sqrt x}-\frac{1}{\sqrt x}\]
is this the function you want to differentiate...
i just noticed pooja195 his answer is correct...
Yes, thanks to everyone for clarifying this one. I think I got it.