यदि y = (sin x)y है तो \(\dfrac{dy}{dx}\) का मान क्या है?

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DSSSB TGT Maths Female Subject Concerned - 22 Sep 2018 Shift 2
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  1. \(\rm \dfrac{y^2\ cot\ x}{1-y\ log(sin\ x)}\)
  2. \(\rm \dfrac{y^2\ cot\ x}{1-y\ log\ x}\)
  3. \(\rm \dfrac{y^2\ cot\ x}{1+y\ log(sin\ x)}\)
  4. \(\rm \dfrac{y^2\ cot\ x}{1+y\ log\ x}\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \dfrac{y^2\ cot\ x}{1-y\ log(sin\ x)}\)
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DSSSB TGT Hindi Female 4th Sep 2021 Shift 2
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200 Questions 200 Marks 120 Mins

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संकल्पना:

दो फलन के गुणनफल का अवकलज पहले फलन के अवकलज को दूसरे फलन से गुणा करके साथ ही पहले फलन को दूसरे फलन के अवकलज से गुणा करना होता है। 

\(f(x) = g(x)h(x) \Rightarrow f'(x) = g'(x)h(x) + g(x)h'(x)\)

गणना:

y = (sin x)y

दोनों पक्षों पर log लेने पर, 

logy = y log(sinx)

दोनों पक्षों में अवकलन करने पर

\(f(x) = g(x)h(x) \Rightarrow f'(x) = g'(x)h(x) + g(x)h'(x)\)

यहाँ, g(x) = y और h(x) = log(sinx)

\(\frac{1}{y} \times \frac{{dy}}{{dx}} = y \times \frac{1}{{\sin x}} \times \cos x + \log \sin x \times \frac{{dy}}{{dx}}\)

\(\frac{{dy}}{{dx}}\left[ {\frac{1}{y} - \log \sin x} \right] = y\cot x\)

\(\frac{{dy}}{{dx}}\left[ {1 - y\log \sin x} \right] = {y^2}\cot x\)

\(\frac{{dy}}{{dx}} = \frac{{{y^2}\cot x}}{{1 - y\log (sin x)}}\)

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