Finding local extrema: Derivative of quotient
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$$f(x) = frac{ln(x)}{sqrt x}$$
I am trying to take the derivative of this.
What I have is:
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}$$
simplifying, I get:
$$-0.5cdotfrac {ln(x)}{x^{3/2}} + x^frac {-1}{2}$$
-- At this point my answer differs from the solution for the term on the right hand side... I get $x^{-1/2}$ and they have $x^{-3/2}$. But I really don't understand how they got that.
And then, if it is correct up to here, I'm not sure how to solve it. If anyone could help me I would really appreciate it. Thanks!
derivatives
edited Nov 18 at 18:27
gimusi
87.4k74393
asked Nov 18 at 18:19
M Do
143
add a comment |
up vote
0
down vote
favorite
$$f(x) = frac{ln(x)}{sqrt x}$$
I am trying to take the derivative of this.
What I have is:
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}$$
simplifying, I get:
$$-0.5cdotfrac {ln(x)}{x^{3/2}} + x^frac {-1}{2}$$
-- At this point my answer differs from the solution for the term on the right hand side... I get $x^{-1/2}$ and they have $x^{-3/2}$. But I really don't understand how they got that.
And then, if it is correct up to here, I'm not sure how to solve it. If anyone could help me I would really appreciate it. Thanks!
derivatives
edited Nov 18 at 18:27
gimusi
87.4k74393
asked Nov 18 at 18:19
M Do
143
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$f(x) = frac{ln(x)}{sqrt x}$$
I am trying to take the derivative of this.
What I have is:
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}$$
simplifying, I get:
$$-0.5cdotfrac {ln(x)}{x^{3/2}} + x^frac {-1}{2}$$
-- At this point my answer differs from the solution for the term on the right hand side... I get $x^{-1/2}$ and they have $x^{-3/2}$. But I really don't understand how they got that.
And then, if it is correct up to here, I'm not sure how to solve it. If anyone could help me I would really appreciate it. Thanks!
derivatives
edited Nov 18 at 18:27
gimusi
87.4k74393
asked Nov 18 at 18:19
M Do
143
$$f(x) = frac{ln(x)}{sqrt x}$$
I am trying to take the derivative of this.
What I have is:
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}$$
simplifying, I get:
$$-0.5cdotfrac {ln(x)}{x^{3/2}} + x^frac {-1}{2}$$
-- At this point my answer differs from the solution for the term on the right hand side... I get $x^{-1/2}$ and they have $x^{-3/2}$. But I really don't understand how they got that.
And then, if it is correct up to here, I'm not sure how to solve it. If anyone could help me I would really appreciate it. Thanks!
derivatives
derivatives
edited Nov 18 at 18:27
gimusi
87.4k74393
asked Nov 18 at 18:19
M Do
143
edited Nov 18 at 18:27
gimusi
87.4k74393
asked Nov 18 at 18:19
M Do
143
edited Nov 18 at 18:27
gimusi
87.4k74393
edited Nov 18 at 18:27
gimusi
87.4k74393
edited Nov 18 at 18:27
gimusi
87.4k74393
87.4k74393
asked Nov 18 at 18:19
M Do
143
asked Nov 18 at 18:19
M Do
143
asked Nov 18 at 18:19
M Do
143
143
add a comment |
add a comment |
1 Answer
1
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0
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We obtain
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}=frac{sqrt x}{x^2}-frac 1{2xsqrt x} ln(x)=frac{1}{xsqrt x}-frac 1{2xsqrt x} ln(x)$$
which agrees with the solution you are looking for.
answered Nov 18 at 18:22
gimusi
87.4k74393
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We obtain
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}=frac{sqrt x}{x^2}-frac 1{2xsqrt x} ln(x)=frac{1}{xsqrt x}-frac 1{2xsqrt x} ln(x)$$
which agrees with the solution you are looking for.
answered Nov 18 at 18:22
gimusi
87.4k74393
add a comment |
up vote
0
down vote
We obtain
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}=frac{sqrt x}{x^2}-frac 1{2xsqrt x} ln(x)=frac{1}{xsqrt x}-frac 1{2xsqrt x} ln(x)$$
which agrees with the solution you are looking for.
answered Nov 18 at 18:22
gimusi
87.4k74393
add a comment |
up vote
0
down vote
up vote
0
down vote
We obtain
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}=frac{sqrt x}{x^2}-frac 1{2xsqrt x} ln(x)=frac{1}{xsqrt x}-frac 1{2xsqrt x} ln(x)$$
which agrees with the solution you are looking for.
answered Nov 18 at 18:22
gimusi
87.4k74393
We obtain
$$f'(x)=frac {frac1x sqrt x - frac 1{2sqrt x} ln(x)}{x}=frac{sqrt x}{x^2}-frac 1{2xsqrt x} ln(x)=frac{1}{xsqrt x}-frac 1{2xsqrt x} ln(x)$$
which agrees with the solution you are looking for.
answered Nov 18 at 18:22
gimusi
87.4k74393
answered Nov 18 at 18:22
gimusi
87.4k74393
answered Nov 18 at 18:22
gimusi
87.4k74393
answered Nov 18 at 18:22
gimusi
87.4k74393
87.4k74393
add a comment |
add a comment |
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