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pooja195
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0That's not the chain rule, that's the power rule.

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

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

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

avendanl
 2 years 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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