Advertisements
Advertisements
प्रश्न
Differentiate \[\left( \cos x \right)^{\sin x }\] with respect to \[\left( \sin x \right)^{\cos x }\]?
उत्तर
\[\text { Let, u } = \left( \cos x \right)^{\sin x} \]
Taking log on both sides,
\[\log u = \log \left( \cos x \right)^{\sin x } \]
\[ \Rightarrow \log u = \sin x \log\left( \cos x \right)\]
Differentiating it with respect to x using chain rule,
\[\frac{1}{u}\frac{du}{dx} = \sin x\frac{d}{dx}\left( \log \cos x \right) + \log \cos x\frac{d}{dx}\left( \sin x \right) \left[ \text{ using product rule } \right]\]
\[ \Rightarrow \frac{1}{u}\frac{du}{dx} = \sin x\left( \frac{1}{\cos x} \right)\frac{d}{dx}\left( \cos x \right) + \log \cos x\left( \cos x \right)\]
\[ \Rightarrow \frac{du}{dx} = u\left[ \left( \tan x \right) \times \left( - \sin x \right) + \log \log x\left( \cos x \right) \right]\]
\[ \Rightarrow \frac{du}{dx} = \left( \cos x \right)^{ \sin x } \left[ \cos x \log\cos x - \sin x \tan x \right] . . . \left( i \right)\]
\[\text { Let, v }= \left( \sin x \right)^{\cos x }\]
Taking log on both sides,
\[\log v = \log \left( \sin x \right)^{\cos x} \]
\[ \Rightarrow \log v = \cos x \log\left( \sin x \right)\]
Differentiating it with respect to x using chain rule,
\[\frac{1}{v}\frac{dv}{dx} = \cos x\frac{d}{dx}\left( \log\sin x \right) + \log\sin x\frac{d}{dx}\left( \cos x \right) ..........\left[ \text { using product rule } \right]\]
\[ \Rightarrow \frac{1}{v}\frac{dv}{dx} = \cos x\left( \frac{1}{\sin x} \right)\frac{d}{dx}\left( \sin x \right) + \log\sin x\left( - \sin x \right)\]
\[ \Rightarrow \frac{dv}{dx} = v\left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]\]
\[ \Rightarrow \frac{dv}{dx} = \left( \sin x \right)^{\cos x } \left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]\]
\[\text { dividing equation }\left( i \right) by \left( ii \right), \]
\[ \therefore \frac{du}{dv} = \frac{\left( \cos x \right)^{\sin x } \left[ \cos x \log\cos x - \sin x \tan x \right]}{\left( \sin x \right)^{\cos x } \left[ \cot x\left( \cos x \right) - \sin x \log\sin x \right]}\]
APPEARS IN
संबंधित प्रश्न
Differentiate the following functions from first principles \[e^\sqrt{2x}\].
Differentiate \[3^{x^2 + 2x}\] ?
Differentiate \[\log \sqrt{\frac{1 - \cos x}{1 + \cos x}}\] ?
Differentiate \[\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}\] ?
Differentiate \[x \sin 2x + 5^x + k^k + \left( \tan^2 x \right)^3\] ?
Differentiate \[\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}\] ?
Differentiate \[\sin^{- 1} \left( \frac{x}{\sqrt{x^2 + a^2}} \right)\] ?
If \[y = \log \left\{ \sqrt{x - 1} - \sqrt{x + 1} \right\}\] ,show that \[\frac{dy}{dx} = \frac{- 1}{2\sqrt{x^2 - 1}}\] ?
If \[y = \frac{x}{x + 2}\] , prove tha \[x\frac{dy}{dx} = \left( 1 - y \right) y\] ?
If \[y = \cos^{- 1} \left\{ \frac{2x - 3 \sqrt{1 - x^2}}{\sqrt{13}} \right\}, \text{ find } \frac{dy}{dx}\] ?
If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?
Differentiate \[\left( \sin x \right)^{\cos x}\] ?
Differentiate \[\sin \left( x^x \right)\] ?
Differentiate \[\left( \cos x \right)^x + \left( \sin x \right)^{1/x}\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
If \[y^x = e^{y - x}\] ,prove that \[\frac{dy}{dx} = \frac{\left( 1 + \log y \right)^2}{\log y}\] ?
If \[\left( \sin x \right)^y = \left( \cos y \right)^x ,\], prove that \[\frac{dy}{dx} = \frac{\log \cos y - y cot x}{\log \sin x + x \tan y}\] ?
If \[y = x \sin y\] , prove that \[\frac{dy}{dx} = \frac{y}{x \left( 1 - x \cos y \right)}\] ?
If \[y = \sqrt{x + \sqrt{x + \sqrt{x + . . . to \infty ,}}}\] prove that \[\frac{dy}{dx} = \frac{1}{2 y - 1}\] ?
If \[\frac{dy}{dx}\] when \[x = a \cos \theta \text{ and } y = b \sin \theta\] ?
Differentiate x2 with respect to x3
Differentiate \[\tan^{- 1} \left( \frac{2x}{1 - x^2} \right)\] with respect to \[\cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right),\text { if }0 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\] with respect to \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right), \text { if } - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}\] ?
If \[f\left( 0 \right) = f\left( 1 \right) = 0, f'\left( 1 \right) = 2 \text { and y } = f \left( e^x \right) e^{f \left( x \right)}\] write the value of \[\frac{dy}{dx} \text{ at x } = 0\] ?
If \[f\left( x \right) = \tan^{- 1} \sqrt{\frac{1 + \sin x}{1 - \sin x}}, 0 \leq x \leq \pi/2, \text{ then } f' \left( \pi/6 \right) \text{ is }\] _________ .
If \[f\left( x \right) = \sqrt{x^2 - 10x + 25}\] then the derivative of f (x) in the interval [0, 7] is ____________ .
If \[y = \log \left( \frac{1 - x^2}{1 + x^2} \right), \text { then } \frac{dy}{dx} =\] __________ .
Find the second order derivatives of the following function log (log x) ?
If \[y = \frac{\log x}{x}\] show that \[\frac{d^2 y}{d x^2} = \frac{2 \log x - 3}{x^3}\] ?
If x = a (θ − sin θ), y = a (1 + cos θ) prove that, find \[\frac{d^2 y}{d x^2}\] ?
If x = cos θ, y = sin3 θ, prove that \[y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 = 3 \sin^2 \theta\left( 5 \cos^2 \theta - 1 \right)\] ?
\[\text { If x } = \cos t + \log \tan\frac{t}{2}, y = \sin t, \text { then find the value of } \frac{d^2 y}{d t^2} \text { and } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
If x = t2 and y = t3, find \[\frac{d^2 y}{d x^2}\] ?
If \[y = \left| \log_e x \right|\] find\[\frac{d^2 y}{d x^2}\] ?
If x = 2 at, y = at2, where a is a constant, then \[\frac{d^2 y}{d x^2} \text { at x } = \frac{1}{2}\] is
If \[\frac{d}{dx}\left[ x^n - a_1 x^{n - 1} + a_2 x^{n - 2} + . . . + \left( - 1 \right)^n a_n \right] e^x = x^n e^x\] then the value of ar, 0 < r ≤ n, is equal to
f(x) = 3x2 + 6x + 8, x ∈ R
