Advertisements
Advertisements
Question
Differentiate the following function from first principles \[e^\sqrt{\cot x}\] .
Advertisements
Solution
\[\text{Let} f\left( x \right) = e^\sqrt{\cot x} \]
\[ \Rightarrow f\left( x + h \right) = e^\sqrt{\cot\left( x + h \right)} \]
\[ \therefore \frac{d}{dx}\left\{ f\left( x \right) \right\} = \lim_{h \to 0} \frac{f\left( x + h \right) - f\left( x \right)}{h}\]
\[ = \lim_{h \to 0} \frac{e^\sqrt{\cot\left( x + h \right)} - e^\sqrt{\cot x}}{h}\]
\[ = \lim_{h \to 0} \frac{e^\sqrt{\cot x} \left( e^{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} - 1 \right)}{h}\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \left(\frac{e^{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} - 1}{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}} \right) \times \left( \frac{\sqrt{\cot\left( x + h \right)} - \sqrt{\cot x}}{h} \right)\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\left( \sqrt{\cot\left( x + h \right)} - \sqrt{\cot x} \right)}{h} \times \frac{\sqrt{\cot\left( x + h \right)} + \sqrt{\cot x}}{\sqrt{\cot\left( x + h \right)} + \sqrt{\cot x}} \left[ \because \lim_{x \to 0} \frac{e^x - 1}{x} = 1 \text{ and rationalizing the numerator } \right]\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right) - \cot x}{h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\frac{\cot\left( x + h \right)\cot x + 1}{\cot\left( x - x - h \right)}}{h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)} \left[ \because \cot\left( A - B \right) = \frac{\cot A\cot B + 1}{\cot B - \cot A} \right]\]
\[ = e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right)\cot x + 1}{\cot\left( - h \right) \times h\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = - e^\sqrt{\cot x} \lim_{h \to 0} \frac{\cot\left( x + h \right)\cot x + 1}{\left( \frac{h}{\tanh} \right)\left( \sqrt{\cot\left( x + h \right)} + \sqrt{\cot x} \right)}\]
\[ = \frac{e^\sqrt{\cot x} \times \left( \cot^2 x + 1 \right)}{2\sqrt{\cot x}} \left[ \because \lim_{x \to 0} \frac{\tan x}{x} = 1 \right]\]
\[ = - \frac{e^\sqrt{\cot x} \times {cosec}^2 x}{2\sqrt{\cot x}} \left[ \because \left( 1 + \cot^2 x \right) = {cosec}^2 x \right]\]
\[ \therefore \frac{d}{dx}\left( e^\sqrt{cot x} \right) = - \frac{e^\sqrt{\cot x} \times {cosec}^2 x}{2\sqrt{\cot x}}\]
APPEARS IN
RELATED QUESTIONS
Prove that `y=(4sintheta)/(2+costheta)-theta `
Differentiate the following functions from first principles log cos x ?
Differentiate the following functions from first principles sin−1 (2x + 3) ?
Differentiate \[\frac{e^x \log x}{x^2}\] ?
Differentiate \[\sqrt{\tan^{- 1} \left( \frac{x}{2} \right)}\] ?
Differentiate \[\log \left( \tan^{- 1} x \right)\]?
Differentiate \[\sin^2 \left\{ \log \left( 2x + 3 \right) \right\}\] ?
If \[y = \frac{x \sin^{- 1} x}{\sqrt{1 - x^2}}\] , prove that \[\left( 1 - x^2 \right) \frac{dy}{dx} = x + \frac{y}{x}\] ?
If \[y = \left( x - 1 \right) \log \left( x - 1 \right) - \left( x + 1 \right) \log \left( x + 1 \right)\] , prove that \[\frac{dy}{dc} = \log \left( \frac{x - 1}{1 + x} \right)\] ?
Differentiate \[\sin^{- 1} \left\{ \frac{x}{\sqrt{x^2 + a^2}} \right\}\] ?
Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{x - a}{x + a} \right)\] ?
Find \[\frac{dy}{dx}\] in the following case \[4x + 3y = \log \left( 4x - 3y \right)\] ?
Differentiate \[\sin \left( x^x \right)\] ?
Differentiate \[x^{\sin^{- 1} x}\] ?
If `y=(sinx)^x + sin^-1 sqrtx "then find" dy/dx`
Find \[\frac{dy}{dx}\] \[y = x^x + \left( \sin x \right)^x\] ?
Find \[\frac{dy}{dx}\]
\[y = x^x + x^{1/x}\] ?
If \[y = \log\frac{x^2 + x + 1}{x^2 - x + 1} + \frac{2}{\sqrt{3}} \tan^{- 1} \left( \frac{\sqrt{3} x}{1 - x^2} \right), \text{ find } \frac{dy}{dx} .\] ?
Find \[\frac{dy}{dx}\] , when \[x = \cos^{- 1} \frac{1}{\sqrt{1 + t^2}} \text{ and y } = \sin^{- 1} \frac{t}{\sqrt{1 + t^2}}, t \in R\] ?
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\[\frac{dy}{dx}\] ?
\[\text { If }x = \cos t\left( 3 - 2 \cos^2 t \right), y = \sin t\left( 3 - 2 \sin^2 t \right) \text { find the value of } \frac{dy}{dx}\text{ at }t = \frac{\pi}{4}\] ?
Differentiate x2 with respect to x3
Differentiate\[\tan^{- 1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right)\] with respect to \[\sin^{-1} \left( \frac{2x}{1 + x^2} \right)\], If \[- 1 < x < 1, x \neq 0 .\] ?
Differentiate \[\sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] 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 \[\frac{\pi}{2} \leq x \leq \frac{3\pi}{2} \text { and y } = \sin^{- 1} \left( \sin x \right), \text { find } \frac{dy}{dx} \] ?
If \[y = \sin^{- 1} \left( \frac{2x}{1 + x^2} \right)\] write the value of \[\frac{dy}{dx}\text { for } x > 1\] ?
If \[y = \sec^{- 1} \left( \frac{x + 1}{x - 1} \right) + \sin^{- 1} \left( \frac{x - 1}{x + 1} \right)\] then write the value of \[\frac{dy}{dx} \] ?
For the curve \[\sqrt{x} + \sqrt{y} = 1, \frac{dy}{dx}\text { at } \left( 1/4, 1/4 \right)\text { is }\] _____________ .
If \[3 \sin \left( xy \right) + 4 \cos \left( xy \right) = 5, \text { then } \frac{dy}{dx} =\] _____________ .
If \[f\left( x \right) = \left| x^2 - 9x + 20 \right|\] then `f' (x)` is equal to ____________ .
If \[f\left( x \right) = \left( \frac{x^l}{x^m} \right)^{l + m} \left( \frac{x^m}{x^n} \right)^{m + n} \left( \frac{x^n}{x^l} \right)^{n + 1}\] the f' (x) is equal to _____________ .
Find the second order derivatives of the following function ex sin 5x ?
If x = a (θ + sin θ), y = a (1 + cos θ), prove that \[\frac{d^2 y}{d x^2} = - \frac{a}{y^2}\] ?
If `x = sin(1/2 log y)` show that (1 − x2)y2 − xy1 − a2y = 0.
\[\text { If }y = A e^{- kt} \cos\left( pt + c \right), \text { prove that } \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 \] ?
If \[y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!}\] .....to ∞, then write \[\frac{d^2 y}{d x^2}\] in terms of y ?
If x = f(t) and y = g(t), then \[\frac{d^2 y}{d x^2}\] is equal to
