Advertisements
Advertisements
प्रश्न
Differentiate \[\log \left( \cos x^2 \right)\] ?
उत्तर
\[\text{Let } y = \log \left( \cos x^2 \right)\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left\{ \log\left( \cos x^2 \right) \right\}\]
\[ = \frac{- 2x \sin x^2}{\cos x^2} \]
\[ = - 2x \tan x^2 \]
\[So, \frac{d}{dx}\left\{ \log\left( \cos x^2 \right) \right\} = - 2x \tan x^2\]
APPEARS IN
संबंधित प्रश्न
If the sum of the lengths of the hypotenuse and a side of a right triangle is given, show that the area of the triangle is maximum, when the angle between them is 60º.
Differentiate the following functions from first principles e−x.
Differentiate the following functions from first principles eax+b.
Differentiate etan x ?
Differentiate tan 5x° ?
Differentiate \[\sin \left( 2 \sin^{- 1} x \right)\] ?
Differentiate \[\frac{2^x \cos x}{\left( x^2 + 3 \right)^2}\] ?
If \[y = \sqrt{x + 1} + \sqrt{x - 1}\] , prove that \[\sqrt{x^2 - 1}\frac{dy}{dx} = \frac{1}{2}y\] ?
Differentiate \[\sin^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x \in R\] ?
Differentiate \[\tan^{- 1} \left( \frac{a + b \tan x}{b - a \tan x} \right)\] ?
Differentiate \[\left( \sin x \right)^{\cos x}\] ?
Differentiate \[{10}^\left( {10}^x \right)\] ?
Differentiate \[x^{\sin^{- 1} x}\] ?
Differentiate \[e^{\sin x }+ \left( \tan x \right)^x\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
Find \[\frac{dy}{dx}\] \[y = x^x + \left( \sin x \right)^x\] ?
If \[y^x = e^{y - x}\] ,prove that \[\frac{dy}{dx} = \frac{\left( 1 + \log y \right)^2}{\log y}\] ?
If \[\left( \cos x \right)^y = \left( \cos y \right)^x , \text{ find } \frac{dy}{dx}\] ?
If \[y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + . . to \infty}}}\] , prove that \[\frac{dy}{dx} = \frac{\sec^2 x}{2 y - 1}\] ?
Find \[\frac{dy}{dx}\] , when \[x = \frac{1 - t^2}{1 + t^2} \text{ and y } = \frac{2 t}{1 + t^2}\] ?
Differentiate x2 with respect to x3
Differentiate \[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cos^{- 1} x, \text { if}\] \[x \in \left( - 1, 0 \right)\] ?
Differentiate \[\sin^{- 1} \left( 2x \sqrt{1 - x^2} \right)\] with respect to \[\sec^{- 1} \left( \frac{1}{\sqrt{1 - x^2}} \right)\], if \[x \in \left( \frac{1}{\sqrt{2}}, 1 \right)\] ?
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 (x) = logx2 (log x), the `f' (x)` at x = e is ____________ .
The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]
For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text { at } \left( 1/4, 1/4 \right)\text { is }\] _____________ .
Let \[\cup = \sin^{- 1} \left( \frac{2 x}{1 + x^2} \right) \text { and }V = \tan^{- 1} \left( \frac{2 x}{1 - x^2} \right), \text { then } \frac{d \cup}{dV} =\] ____________ .
If \[\sin y = x \sin \left( a + y \right), \text { then }\frac{dy}{dx} \text { is}\] ____________ .
If \[y = \log \sqrt{\tan x}\] then the value of \[\frac{dy}{dx}\text { at }x = \frac{\pi}{4}\] is given by __________ .
Find the second order derivatives of the following function log (sin x) ?
If y = ex (sin x + cos x) prove that \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\] ?
If y = sin (log x), prove that \[x^2 \frac{d^2 y}{d x^2} + x\frac{dy}{dx} + y = 0\] ?
\[\text { If x } = \cos t + \log \tan\frac{t}{2}, y = \sin t, \text { then find the value of } \frac{d^2 y}{d t^2} \text { and } \frac{d^2 y}{d x^2} \text { at } t = \frac{\pi}{4} \] ?
\[\text { If y } = a \left\{ x + \sqrt{x^2 + 1} \right\}^n + b \left\{ x - \sqrt{x^2 + 1} \right\}^{- n} , \text { prove that }\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \]
Disclaimer: There is a misprint in the question,
\[\left( x^2 + 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0\] must be written instead of
\[\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + x\frac{d y}{d x} - n^2 y = 0 \] ?
If x = f(t) and y = g(t), then write the value of \[\frac{d^2 y}{d x^2}\] ?
If y = a + bx2, a, b arbitrary constants, then
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
