Find the value of differentiated function
$begingroup$
Let $f(x) = cos left[cot ^{-1}left( frac{cos x}{sqrt {1 - cos 2x}}right)right]$ where $frac{pi}{4} < x < frac{pi}{2}$, then what is the value of $ frac{d(f(x))}{d(cot(x))}$?
MY try :
enter image description here
Answer is given as 1
functions trigonometry
asked Dec 12 '18 at 5:26
faceface
184
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add a comment |
$begingroup$
Let $f(x) = cos left[cot ^{-1}left( frac{cos x}{sqrt {1 - cos 2x}}right)right]$ where $frac{pi}{4} < x < frac{pi}{2}$, then what is the value of $ frac{d(f(x))}{d(cot(x))}$?
MY try :
enter image description here
Answer is given as 1
functions trigonometry
asked Dec 12 '18 at 5:26
faceface
184
$endgroup$
add a comment |
$begingroup$
Let $f(x) = cos left[cot ^{-1}left( frac{cos x}{sqrt {1 - cos 2x}}right)right]$ where $frac{pi}{4} < x < frac{pi}{2}$, then what is the value of $ frac{d(f(x))}{d(cot(x))}$?
MY try :
enter image description here
Answer is given as 1
functions trigonometry
asked Dec 12 '18 at 5:26
faceface
184
$endgroup$
Let $f(x) = cos left[cot ^{-1}left( frac{cos x}{sqrt {1 - cos 2x}}right)right]$ where $frac{pi}{4} < x < frac{pi}{2}$, then what is the value of $ frac{d(f(x))}{d(cot(x))}$?
MY try :
enter image description here
Answer is given as 1
functions trigonometry
functions trigonometry
asked Dec 12 '18 at 5:26
faceface
184
asked Dec 12 '18 at 5:26
faceface
184
edited Dec 12 '18 at 10:08
face
asked Dec 12 '18 at 5:26
faceface
184
asked Dec 12 '18 at 5:26
faceface
184
asked Dec 12 '18 at 5:26
faceface
184
184
add a comment |
add a comment |
1 Answer
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$begingroup$
You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cosBigg(cot^{-1}Big(frac{cot(x)}{sqrt{2}}Big)Bigg)$$
(Note that I have corrected your calculation mistake and written $sqrt{2}$ instead of $2$)
Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(frac{y}{sqrt{2}}))$. Apply the identity,
$$cos(cot^{-1}(x))=frac{x}{sqrt{x^2+1}}$$
Now the function reduces to $frac{y}{sqrt{y^2+2}}$ and you can differentiate it easily.
Hope it helps:)
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
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1 Answer
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$begingroup$
You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cosBigg(cot^{-1}Big(frac{cot(x)}{sqrt{2}}Big)Bigg)$$
(Note that I have corrected your calculation mistake and written $sqrt{2}$ instead of $2$)
Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(frac{y}{sqrt{2}}))$. Apply the identity,
$$cos(cot^{-1}(x))=frac{x}{sqrt{x^2+1}}$$
Now the function reduces to $frac{y}{sqrt{y^2+2}}$ and you can differentiate it easily.
Hope it helps:)
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
$endgroup$
add a comment |
$begingroup$
You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cosBigg(cot^{-1}Big(frac{cot(x)}{sqrt{2}}Big)Bigg)$$
(Note that I have corrected your calculation mistake and written $sqrt{2}$ instead of $2$)
Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(frac{y}{sqrt{2}}))$. Apply the identity,
$$cos(cot^{-1}(x))=frac{x}{sqrt{x^2+1}}$$
Now the function reduces to $frac{y}{sqrt{y^2+2}}$ and you can differentiate it easily.
Hope it helps:)
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
$endgroup$
add a comment |
$begingroup$
You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cosBigg(cot^{-1}Big(frac{cot(x)}{sqrt{2}}Big)Bigg)$$
(Note that I have corrected your calculation mistake and written $sqrt{2}$ instead of $2$)
Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(frac{y}{sqrt{2}}))$. Apply the identity,
$$cos(cot^{-1}(x))=frac{x}{sqrt{x^2+1}}$$
Now the function reduces to $frac{y}{sqrt{y^2+2}}$ and you can differentiate it easily.
Hope it helps:)
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
$endgroup$
You need not apply the chain rule. As you have already solved upto the following step,$$f(x)=cosBigg(cot^{-1}Big(frac{cot(x)}{sqrt{2}}Big)Bigg)$$
(Note that I have corrected your calculation mistake and written $sqrt{2}$ instead of $2$)
Now, let cot$x$=$y$, given function becomes, $cos(cot^{-1}(frac{y}{sqrt{2}}))$. Apply the identity,
$$cos(cot^{-1}(x))=frac{x}{sqrt{x^2+1}}$$
Now the function reduces to $frac{y}{sqrt{y^2+2}}$ and you can differentiate it easily.
Hope it helps:)
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
answered Dec 12 '18 at 10:48
MartundMartund
1,645213
1,645213
add a comment |
add a comment |
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