Advertisements
Advertisements
प्रश्न
Differentiate the following functions from first principles \[e^\sqrt{2x}\].
उत्तर
\[\text{ Let } f\left( x \right) = e^\sqrt{2x} \]
\[ \Rightarrow f\left( x + h \right) = e^\sqrt{2\left( x + h \right)} \]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^{{}^\sqrt{2\left( x + h \right)}} - e^{{}^\sqrt{2x}}}{h}\]
\[ = \lim_{h \to 0} e^\sqrt{2x} \left[ \frac{e^\sqrt{2\left( x + h \right)} - \sqrt{2x} - 1}{h} \right]\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \left[ \frac{e^\sqrt{2\left( x + h \right)} - \sqrt{2x} - 1}{\sqrt{2\left( x + h \right)} - \sqrt{2x}} \right] \times \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h} \left[ \because \lim_{h \to 0} \frac{e^h - 1}{h} = 1 \right]\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{\sqrt{2\left( x + h \right)} - \sqrt{2x}}{h} \times \frac{\sqrt{2\left( x + h \right)} + \sqrt{2x}}{\sqrt{2\left( x + h \right)} + \sqrt{2x}} \] [Rationalising the numerator]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2\left( x + h \right) - 2x}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2x + 2h - 2x}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)} \]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2h}{h\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = e^\sqrt{2x} \lim_{h \to 0} \frac{2}{\left( \sqrt{2\left( x + h \right)} + \sqrt{2x} \right)}\]
\[ = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]
\[\text{ Hence }, \frac{d}{dx}\left( e^\sqrt{2x} \right) = \frac{e^\sqrt{2x}}{\sqrt{2x}}\]
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 function from first principles \[e^\sqrt{\cot x}\] .
Differentiate tan (x° + 45°) ?
Differentiate log7 (2x − 3) ?
Differentiate \[3^{x^2 + 2x}\] ?
Differentiate \[e^{ax} \sec x \tan 2x\] ?
If xy = 4, prove that \[x\left( \frac{dy}{dx} + y^2 \right) = 3 y\] ?
Differentiate \[\tan^{- 1} \left( \frac{x}{1 + 6 x^2} \right)\] ?
Differentiate
\[\tan^{- 1} \left( \frac{\cos x + \sin x}{\cos x - \sin x} \right), \frac{\pi}{4} < x < \frac{\pi}{4}\] ?
If \[y = se c^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right), x > 0 . \text{ Find} \frac{dy}{dx}\] ?
If \[\log \sqrt{x^2 + y^2} = \tan^{- 1} \left( \frac{y}{x} \right)\] Prove that \[\frac{dy}{dx} = \frac{x + y}{x - y}\] ?
If \[x \sin \left( a + y \right) + \sin a \cos \left( a + y \right) = 0\] Prove that \[\frac{dy}{dx} = \frac{\sin^2 \left( a + y \right)}{\sin a}\] ?
If \[y = x \sin y\] , Prove that \[\frac{dy}{dx} = \frac{\sin y}{\left( 1 - x \cos y \right)}\] ?
If \[\tan \left( x + y \right) + \tan \left( x - y \right) = 1, \text{ find} \frac{dy}{dx}\] ?
If \[\cos y = x \cos \left( a + y \right), \text{ with } \cos a \neq \pm 1, \text{ prove that } \frac{dy}{dx} = \frac{\cos^2 \left( a + y \right)}{\sin a}\] ?
Differentiate \[\left( 1 + \cos x \right)^x\] ?
Differentiate \[\left( x \cos x \right)^x + \left( x \sin x \right)^{1/x}\] ?
If \[y = \sin \left( x^x \right)\] prove that \[\frac{dy}{dx} = \cos \left( x^x \right) \cdot x^x \left( 1 + \log x \right)\] ?
If \[x^y + y^x = \left( x + y \right)^{x + y} , \text{ find } \frac{dy}{dx}\] ?
If \[y = \sqrt{x + \sqrt{x + \sqrt{x + . . . to \infty ,}}}\] prove that \[\frac{dy}{dx} = \frac{1}{2 y - 1}\] ?
Find \[\frac{dy}{dx}\] ,when \[x = \frac{e^t + e^{- t}}{2} \text{ and } y = \frac{e^t - e^{- t}}{2}\] ?
If \[x = e^{\cos 2 t} \text{ and y }= e^{\sin 2 t} ,\] prove that \[\frac{dy}{dx} = - \frac{y \log x}{x \log y}\] ?
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 \[y = \sin^{- 1} \left( \sin x \right), - \frac{\pi}{2} \leq x \leq \frac{\pi}{2}\] ,Then, write the value of \[\frac{dy}{dx} \text{ for } x \in \left( - \frac{\pi}{2}, \frac{\pi}{2} \right) \] ?
If \[y = \sin^{- 1} x + \cos^{- 1} x\] ,find \[\frac{dy}{dx}\] ?
If f (x) is an even function, then write whether `f' (x)` is even or odd ?
For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text { at } \left( 1/4, 1/4 \right)\text { is }\] _____________ .
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 = sin t, y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?
If y = (tan−1 x)2, then prove that (1 + x2)2 y2 + 2x(1 + x2)y1 = 2 ?
If y log (1 + cos x), prove that \[\frac{d^3 y}{d x^3} + \frac{d^2 y}{d x^2} \cdot \frac{dy}{dx} = 0\] ?
\[\text { If x } = a \sin t - b \cos t, y = a \cos t + b \sin t, \text { prove that } \frac{d^2 y}{d x^2} = - \frac{x^2 + y^2}{y^3} \] ?
If y = sin (m sin−1 x), then (1 − x2) y2 − xy1 is equal to
If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is
\[\text { If } y = \left( x + \sqrt{1 + x^2} \right)^n , \text { then show that }\]
\[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = n^2 y .\]
If y = xx, prove that \[\frac{d^2 y}{d x^2} - \frac{1}{y} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x} = 0 .\]
If x = a (1 + cos θ), y = a(θ + sin θ), prove that \[\frac{d^2 y}{d x^2} = \frac{- 1}{a}at \theta = \frac{\pi}{2}\]
