Derivative of $ y = frac{1}{ln^{2}x} $ [on hold]
up vote
-2
down vote
favorite
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
put on hold as off-topic by amWhy, DRF, RRL, Did, user302797 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, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
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
put on hold as off-topic by amWhy, DRF, RRL, Did, user302797 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, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
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
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
116k637108
116k637108
asked Nov 25 at 14:04
Johny547
1164
1164
put on hold as off-topic by amWhy, DRF, RRL, Did, user302797 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, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by amWhy, DRF, RRL, Did, user302797 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, user302797
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
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:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
add a comment |
up vote
0
down vote
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
add a comment |
up vote
0
down vote
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
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:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
add a comment |
up vote
2
down vote
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:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
add a comment |
up vote
2
down vote
up vote
2
down vote
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:
$$ frac{mathrm d}{mathrm dx} frac{1}{(ln x)^2} = -frac{2}{(ln x)^3}frac{mathrm d}{mathrm dx}ln x = -frac{2}{x(ln x)^3}. $$
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
5,4591623
5,4591623
add a comment |
add a comment |
up vote
0
down vote
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
add a comment |
up vote
0
down vote
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint:
Use the chain rule, and remember that
$$biggl(frac1{x^2}biggr)'=-frac 2{x^3},enspacetext{more generally: }quadbiggl(frac1{x^n}biggr)'=-frac n{x^{n+1}}$$
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
116k637108
116k637108
add a comment |
add a comment |
up vote
0
down vote
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
add a comment |
up vote
0
down vote
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
add a comment |
up vote
0
down vote
up vote
0
down vote
A rule that always hold is $(u^n)' = nu'u^{n-1}$.
With $u = ln$ we get $forall xin mathbb{R}^*_+, u^{-2}(x) = frac{-2}{x}ln^{-3}(x)$.
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
4229
4229
add a comment |
add a comment |