anonymous
  • anonymous
Differentiate: f(x)=sqrt(x) lnx
Calculus1
chestercat
  • chestercat
See more answers at brainly.com
chestercat
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
Use product rule.
pooja195
  • 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
anonymous
  • anonymous
That's not the chain rule, that's the power rule.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Ahhhh THank You.
pooja195
  • pooja195
so im right
anonymous
  • anonymous
Anything can be written as a chain rule. \[\frac{ d }{ dx } x ^{2}=\frac{ d }{ dx }(x)^{2}=2(x)\frac{ d }{ dx }(x)=2x \times1=2x\]
anonymous
  • anonymous
Well, not everything, but the point is there.
anonymous
  • anonymous
\[\frac{d(\sqrt{x}\ln(x))}{dx}=\frac{1}{2}\frac{\ln(x)}{\sqrt x}-\frac{1}{\sqrt x}\]
anonymous
  • anonymous
is this the function you want to differentiate...
anonymous
  • anonymous
i just noticed pooja195 his answer is correct...
anonymous
  • anonymous
Yes, thanks to everyone for clarifying this one. I think I got it.

Looking for something else?

Not the answer you are looking for? Search for more explanations.