Derivative of $ y = frac{1}{ln^{2}x} $ [closed]
I am supposed to find the derivative of $ y = frac{1}{ln^{2}x} $. How would you calculate it? My first step was to do this: $frac{-1ast ln^{2}x}{(ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ ln^{2}x $. Thanks
derivatives
edited Nov 25 at 14:10
Bernard
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asked Nov 25 at 14:04
Johny547
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closed as off-topic by amWhy, DRF, RRL, Did, Saad Dec 3 at 1:04
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I am supposed to find the derivative of $ y = frac{1}{ln^{2}x} $. How would you calculate it? My first step was to do this: $frac{-1ast ln^{2}x}{(ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ ln^{2}x $. Thanks
derivatives
edited Nov 25 at 14:10
Bernard
117k637111
asked Nov 25 at 14:04
Johny547
1154
closed as off-topic by amWhy, DRF, RRL, Did, Saad Dec 3 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, DRF, RRL, Did, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I am supposed to find the derivative of $ y = frac{1}{ln^{2}x} $. How would you calculate it? My first step was to do this: $frac{-1ast ln^{2}x}{(ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ ln^{2}x $. Thanks
derivatives
edited Nov 25 at 14:10
Bernard
117k637111
asked Nov 25 at 14:04
Johny547
1154
I am supposed to find the derivative of $ y = frac{1}{ln^{2}x} $. How would you calculate it? My first step was to do this: $frac{-1ast ln^{2}x}{(ln^{2}x)^{2}}$. How would you continue? I don't know what to do with $ ln^{2}x $. Thanks
derivatives
derivatives
edited Nov 25 at 14:10
Bernard
117k637111
asked Nov 25 at 14:04
Johny547
1154
edited Nov 25 at 14:10
Bernard
117k637111
asked Nov 25 at 14:04
Johny547
1154
edited Nov 25 at 14:10
Bernard
117k637111
edited Nov 25 at 14:10
Bernard
117k637111
edited Nov 25 at 14:10
Bernard
117k637111
117k637111
asked Nov 25 at 14:04
Johny547
1154
asked Nov 25 at 14:04
Johny547
1154
asked Nov 25 at 14:04
Johny547
1154
1154
closed as off-topic by amWhy, DRF, RRL, Did, Saad Dec 3 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, DRF, RRL, Did, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, DRF, RRL, Did, Saad Dec 3 at 1:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, DRF, RRL, Did, Saad
If this question can be reworded to fit the rules in the help center, please edit the question.
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3 Answers
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You are differentiating
$$ frac{1}{(ln x)^2} = f(g(x)), $$
where
$$ f(x) = frac{1}{x^2}, quad g(x) = ln x. $$
Since
$$ f'(x) = -frac{2}{x^3}, quad g'(x) = frac{1}{x}, $$
the chain rule therefore gives
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = frac{mathrm d}{mathrm dx} f(g(x)) = f'(g(x))g'(x) = -frac{2}{(ln x)^3} frac{1}{x}. $$
You could do it even more directly:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
answered Nov 25 at 14:07
MisterRiemann
5,7291624
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Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
answered Nov 25 at 14:09
Bernard
117k637111
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A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
answered Nov 25 at 14:13
Euler Pythagoras
4949
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are differentiating
$$ frac{1}{(ln x)^2} = f(g(x)), $$
where
$$ f(x) = frac{1}{x^2}, quad g(x) = ln x. $$
Since
$$ f'(x) = -frac{2}{x^3}, quad g'(x) = frac{1}{x}, $$
the chain rule therefore gives
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = frac{mathrm d}{mathrm dx} f(g(x)) = f'(g(x))g'(x) = -frac{2}{(ln x)^3} frac{1}{x}. $$
You could do it even more directly:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
answered Nov 25 at 14:07
MisterRiemann
5,7291624
add a comment |
You are differentiating
$$ frac{1}{(ln x)^2} = f(g(x)), $$
where
$$ f(x) = frac{1}{x^2}, quad g(x) = ln x. $$
Since
$$ f'(x) = -frac{2}{x^3}, quad g'(x) = frac{1}{x}, $$
the chain rule therefore gives
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = frac{mathrm d}{mathrm dx} f(g(x)) = f'(g(x))g'(x) = -frac{2}{(ln x)^3} frac{1}{x}. $$
You could do it even more directly:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
answered Nov 25 at 14:07
MisterRiemann
5,7291624
add a comment |
You are differentiating
$$ frac{1}{(ln x)^2} = f(g(x)), $$
where
$$ f(x) = frac{1}{x^2}, quad g(x) = ln x. $$
Since
$$ f'(x) = -frac{2}{x^3}, quad g'(x) = frac{1}{x}, $$
the chain rule therefore gives
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = frac{mathrm d}{mathrm dx} f(g(x)) = f'(g(x))g'(x) = -frac{2}{(ln x)^3} frac{1}{x}. $$
You could do it even more directly:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
answered Nov 25 at 14:07
MisterRiemann
5,7291624
You are differentiating
$$ frac{1}{(ln x)^2} = f(g(x)), $$
where
$$ f(x) = frac{1}{x^2}, quad g(x) = ln x. $$
Since
$$ f'(x) = -frac{2}{x^3}, quad g'(x) = frac{1}{x}, $$
the chain rule therefore gives
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = frac{mathrm d}{mathrm dx} f(g(x)) = f'(g(x))g'(x) = -frac{2}{(ln x)^3} frac{1}{x}. $$
You could do it even more directly:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
answered Nov 25 at 14:07
MisterRiemann
5,7291624
answered Nov 25 at 14:07
MisterRiemann
5,7291624
answered Nov 25 at 14:07
MisterRiemann
5,7291624
answered Nov 25 at 14:07
MisterRiemann
5,7291624
5,7291624
add a comment |
add a comment |
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
answered Nov 25 at 14:09
Bernard
117k637111
add a comment |
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
answered Nov 25 at 14:09
Bernard
117k637111
add a comment |
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
answered Nov 25 at 14:09
Bernard
117k637111
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
answered Nov 25 at 14:09
Bernard
117k637111
answered Nov 25 at 14:09
Bernard
117k637111
answered Nov 25 at 14:09
Bernard
117k637111
answered Nov 25 at 14:09
Bernard
117k637111
117k637111
add a comment |
add a comment |
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
answered Nov 25 at 14:13
Euler Pythagoras
4949
add a comment |
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
answered Nov 25 at 14:13
Euler Pythagoras
4949
add a comment |
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
answered Nov 25 at 14:13
Euler Pythagoras
4949
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
answered Nov 25 at 14:13
Euler Pythagoras
4949
answered Nov 25 at 14:13
Euler Pythagoras
4949
answered Nov 25 at 14:13
Euler Pythagoras
4949
answered Nov 25 at 14:13
Euler Pythagoras
4949
4949
add a comment |
add a comment |
