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pooja195
 one year ago
Best ResponseYou've already chosen the best response.1Use 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

wio
 one year ago
Best ResponseYou've already chosen the best response.0That's not the chain rule, that's the power rule.

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Anything can be written as a chain rule. \[\frac{ d }{ dx } x ^{2}=\frac{ d }{ dx }(x)^{2}=2(x)\frac{ d }{ dx }(x)=2x \times1=2x\]

calmat01
 one year ago
Best ResponseYou've already chosen the best response.0Well, not everything, but the point is there.

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{d(\sqrt{x}\ln(x))}{dx}=\frac{1}{2}\frac{\ln(x)}{\sqrt x}\frac{1}{\sqrt x}\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0is this the function you want to differentiate...

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0i just noticed pooja195 his answer is correct...

avendanl
 one year ago
Best ResponseYou've already chosen the best response.0Yes, thanks to everyone for clarifying this one. I think I got it.
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