Advertisements
Advertisements
प्रश्न
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 \[\frac{dy}{dx} \text{ at x } = 0\] ?
Advertisements
उत्तर
\[\text{ We have,} f\left( 0 \right) = f\left( 1 \right) = 0 , f'\left( 1 \right) = 2\]
\[\text { and, } \]
\[y = f\left( e^x \right) e^{f\left( x \right)}\]
\[\Rightarrow \frac{dy}{dx} = \frac{d}{dx}\left[ f\left( e^x \right) \times e^{f\left( x \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right)\frac{d}{dx} e^{f\left( x \right)} + e^{f\left( x \right)} \frac{d}{dx}f\left( e^x \right) \left[ \text{Using product rule } \right]\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right) \times e^{f\left( x \right)} \frac{d}{dx}f\left( x \right) + e^{f\left( x \right)} \times f'\left( e^x \right)\frac{d}{dx}\left( e^x \right)\]
\[ \Rightarrow \frac{dy}{dx} = f\left( e^x \right) \times e^{f\left( x \right)} \times f'\left( x \right) + e^{f\left( x \right)} \times f'\left( e^x \right) \times e^x \]
\[\text{ Putting x } = 0, \text{ we get }, \]
\[\frac{dy}{dx} = f\left( e^0 \right) \times e^{f\left( 0 \right)} \times f'\left( 0 \right) + e^{f\left( 0 \right)} \times f'\left( e^0 \right) \times e^0 \]
\[ \Rightarrow \frac{dy}{dx} = f\left( 1 \right) e^{f\left( 0 \right)} \times f'\left( 0 \right) + e^{f\left( 0 \right)} \times f'\left( 1 \right) \times 1\]
\[ \Rightarrow \frac{dy}{dx} = 0 \times e^0 \times f'\left( 0 \right) + e^0 \times 2 \times 1 .........\left[ \because f\left( x \right) = f\left( 1 \right) = 0 \text{ and }f'\left( 1 \right) = 2 \right]\]
\[ \Rightarrow \frac{dy}{dx} = 0 + 1 \times 2 \times 1\]
\[ \Rightarrow \frac{dy}{dx} = 2\]
APPEARS IN
संबंधित प्रश्न
If y = xx, prove that `(d^2y)/(dx^2)−1/y(dy/dx)^2−y/x=0.`
Differentiate the following functions from first principles eax+b.
Differentiate the following functions from first principles log cos x ?
Differentiate sin2 (2x + 1) ?
Differentiate \[\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)\] ?
Differentiate \[\cos^{- 1} \left\{ \sqrt{\frac{1 + x}{2}} \right\}, - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left\{ \frac{x}{\sqrt{a^2 - x^2}} \right\}, - a < x < a\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sqrt{x} + \sqrt{a}}{1 - \sqrt{xa}} \right)\] ?
If \[y = \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) + \sec^{- 1} \left( \frac{1 + x^2}{1 - x^2} \right), x > 0\] ,prove that \[\frac{dy}{dx} = \frac{4}{1 + x^2} \] ?
If \[x \sqrt{1 + y} + y \sqrt{1 + x} = 0\] , prove that \[\left( 1 + x \right)^2 \frac{dy}{dx} + 1 = 0\] ?
Differentiate \[\left( \log x \right)^{\cos x}\] ?
Differentiate\[\left( x + \frac{1}{x} \right)^x + x^\left( 1 + \frac{1}{x} \right)\] ?
Differentiate \[x^{x^2 - 3} + \left( x - 3 \right)^{x^2}\] ?
Find \[\frac{dy}{dx}\] \[y = \frac{e^{ax} \cdot \sec x \cdot \log x}{\sqrt{1 - 2x}}\] ?
Find \[\frac{dy}{dx}\] ,When \[x = a \left( 1 - \cos \theta \right) \text{ and } y = a \left( \theta + \sin \theta \right) \text{ at } \theta = \frac{\pi}{2}\] ?
Find \[\frac{dy}{dx}\] , when \[x = \frac{3 at}{1 + t^2}, \text{ and } y = \frac{3 a t^2}{1 + t^2}\] ?
If \[x = \frac{\sin^3 t}{\sqrt{\cos 2 t}}, y = \frac{\cos^3 t}{\sqrt{\cos t 2 t}}\] , find\[\frac{dy}{dx}\] ?
If \[x = \left( t + \frac{1}{t} \right)^a , y = a^{t + \frac{1}{t}} , \text{ find } \frac{dy}{dx}\] ?
Differentiate \[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cos^{- 1} x, \text { if}\] \[x \in \left( - 1, 0 \right)\] ?
If \[\frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \text { and y } = \sin^{- 1} \left( \sin x \right), \text { find } \frac{dy}{dx} \] ?
If \[- \frac{\pi}{2} < x < 0 \text{ and y } = \tan^{- 1} \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}, \text{ find } \frac{dy}{dx}\] ?
The derivative of \[\sec^{- 1} \left( \frac{1}{2 x^2 + 1} \right) \text { w . r . t }. \sqrt{1 + 3 x} \text { at } x = - 1/3\]
If \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .
If \[\sin^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = \text { log a then } \frac{dy}{dx}\] is equal to _____________ .
Find the second order derivatives of the following function x3 + tan x ?
If y = log (sin x), prove that \[\frac{d^3 y}{d x^3} = 2 \cos \ x \ {cosec}^3 x\] ?
If x = a(1 − cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{1}{a}\text { at } \theta = \frac{\pi}{2}\] ?
If log y = tan−1 x, show that (1 + x2)y2 + (2x − 1) y1 = 0 ?
If y = tan−1 x, show that \[\left( 1 + x^2 \right) \frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\] ?
\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text { and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?
\[\text { If y } = x^n \left\{ a \cos\left( \log x \right) + b \sin\left( \log x \right) \right\}, \text { prove that } x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0 \] Disclaimer: There is a misprint in the question. It must be
\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)x\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] instead of 1
\[x^2 \frac{d^2 y}{d x^2} + \left( 1 - 2n \right)\frac{d y}{d x} + \left( 1 + n^2 \right)y = 0\] ?
If y = a cos (loge x) + b sin (loge x), then x2 y2 + xy1 =
If logy = tan–1 x, then show that `(1+x^2) (d^2y)/(dx^2) + (2x - 1) dy/dx = 0 .`
f(x) = 3x2 + 6x + 8, x ∈ R
Range of 'a' for which x3 – 12x + [a] = 0 has exactly one real root is (–∞, p) ∪ [q, ∞), then ||p| – |q|| is ______.
