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pooja195Best ResponseYou've already chosen the best response.1
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
 one year ago

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

calmat01Best ResponseYou've already chosen the best response.0
Anything can be written as a chain rule. \[\frac{ d }{ dx } x ^{2}=\frac{ d }{ dx }(x)^{2}=2(x)\frac{ d }{ dx }(x)=2x \times1=2x\]
 one year ago

calmat01Best ResponseYou've already chosen the best response.0
Well, not everything, but the point is there.
 one year ago

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

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

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

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