Advertisements
Advertisements
Question
If y `sqrt(x^2 + 1) = log sqrt(x^2 + 1) - x`, show that `(x^2 + 1)(dy)/(dx) + xy + 1 = 0.`
Advertisements
Solution
Given:
y `sqrt(x^2 + 1) = log (sqrt(x^2 + 1) - x)`
Differentiate the Left-Hand Side:
Using the product rule (uv)′ = u′v + uv′:
Let u = y and v = `sqrt(x^2 + 1)`
`d/dx (y sqrt(x^2 + 1)) = (dy)/(dx) . sqrt(x^2 + 1) + y . d/dx (sqrt(x^2 + 1))`
= `sqrt(x^2 + 1) (dy)/(dx) + y . (1/(2sqrt(x^2 + 1)) . 2x)`
= `sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1)` ...(i)
Differentiate the Right-Hand Side:
Using the chain rule for log(u):
`d/dx [log (sqrt(x^2 + 1) - x)] = 1/(sqrt(x^2 + 1) - x) . d/dx (sqrt(x^2 + 1) - x)`
= `1/(sqrt(x^2 + 1) - x) . (x/sqrt(x^2 + 1) - 1)`
Take the LCM in the bracket:
= `1/(sqrt(x^2 + 1) - x) . ((x - sqrt(x^2 + 1))/sqrt(x^2 + 1))`
= `1/(sqrt(x^2 + 1) - x) . ((-sqrt(x^2 + 1) - x)/sqrt(x^2 + 1))`
= `-1/(sqrt(x^2 + 1)` ...(ii)
Equate LHS and RHS
`sqrt(x^2 + 1) (dy)/(dx) + (xy)/sqrt(x^2 + 1) = -1/(sqrt(x^2 + 1)`
Multiply the entire equation by `sqrt(x^2 + 1)` to clear the denominators:
`(sqrt(x^2 + 1) . sqrt(x^2 + 1)) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy = -1`
`(x^2 + 1) (dy)/(dx) + xy + 1 = 0`
Hence proved
APPEARS IN
RELATED QUESTIONS
Differentiate the following functions from first principles \[e^\sqrt{2x}\].
Differentiate the following function from first principles \[e^\sqrt{\cot x}\] .
Differentiate sin (log x) ?
Differentiate `2^(x^3)` ?
Differentiate logx 3 ?
Differentiate \[e^{3 x} \cos 2x\] ?
Differentiate \[e^{\tan^{- 1}} \sqrt{x}\] ?
Differentiate \[\log \left( 3x + 2 \right) - x^2 \log \left( 2x - 1 \right)\] ?
Differentiate \[\frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}\] ?
Differentiate \[\left( \sin^{- 1} x^4 \right)^4\] ?
Differentiate \[\frac{x^2 + 2}{\sqrt{\cos x}}\] ?
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 \[y = e^x + e^{- x}\] prove that \[\frac{dy}{dx} = \sqrt{y^2 - 4}\] ?
Differentiate \[\sin^{- 1} \left\{ \sqrt{\frac{1 - x}{2}} \right\}, 0 < x < 1\] ?
Differentiate \[\sin^{- 1} \left( \frac{x + \sqrt{1 - x^2}}{\sqrt{2}} \right), - 1 < x < 1\] ?
Differentiate \[\tan^{- 1} \left( \frac{\sin x}{1 + \cos x} \right), - \pi < x < \pi\] ?
Differentiate \[\sin^{- 1} \left( \frac{1}{\sqrt{1 + x^2}} \right)\] ?
Differentiate the following with respect to x:
\[\cos^{- 1} \left( \sin x \right)\]
If \[y = \cos^{- 1} \left( 2x \right) + 2 \cos^{- 1} \sqrt{1 - 4 x^2}, 0 < x < \frac{1}{2}, \text{ find } \frac{dy}{dx} .\] ?
Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left(36 \right)^x} \right\}\] with respect to x.
Find \[\frac{dy}{dx}\] in the following case: \[y^3 - 3x y^2 = x^3 + 3 x^2 y\] ?
If \[y = \left\{ \log_{\cos x} \sin x \right\} \left\{ \log_{\sin x} \cos x \right\}^{- 1} + \sin^{- 1} \left( \frac{2x}{1 + x^2} \right), \text{ find } \frac{dy}{dx} \text{ at }x = \frac{\pi}{4}\] ?
If \[\sqrt{y + x} + \sqrt{y - x} = c, \text {show that } \frac{dy}{dx} = \frac{y}{x} - \sqrt{\frac{y^2}{x^2} - 1}\] ?
Differentiate \[x^{\cos^{- 1} x}\] ?
Differentiate \[x^{\tan^{- 1} x }\] ?
Differentiate \[x^\left( \sin x - \cos x \right) + \frac{x^2 - 1}{x^2 + 1}\] ?
Differentiate \[x^{x \cos x +} \frac{x^2 + 1}{x^2 - 1}\] ?
Find \[\frac{dy}{dx}\] \[y = x^x + \left( \sin x \right)^x\] ?
If \[y = x \sin y\] , prove that \[\frac{dy}{dx} = \frac{y}{x \left( 1 - x \cos y \right)}\] ?
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 \[\tan^{- 1} \left( \frac{\cos x}{1 + \sin x} \right)\] with respect to \[\sec^{- 1} x\] ?
If \[x = a \left( \theta + \sin \theta \right), y = a \left( 1 + \cos \theta \right), \text{ find} \frac{dy}{dx}\] ?
If \[y = \tan^{- 1} \left( \frac{1 - x}{1 + x} \right), \text{ find} \frac{dy}{dx}\] ?
If \[y = \log \sqrt{\tan x}, \text{ write } \frac{dy}{dx} \] ?
Given \[f\left( x \right) = 4 x^8 , \text { then }\] _________________ .
If x = a sec θ, y = b tan θ, prove that \[\frac{d^2 y}{d x^2} = - \frac{b^4}{a^2 y^3}\] ?
If x = sin t, y = sin pt, prove that \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?
\[\text { If x } = a\left( \cos2t + 2t \sin2t \right)\text { and y } = a\left( \sin2t - 2t \cos2t \right), \text { then find } \frac{d^2 y}{d x^2} \] ?
If y = axn+1 + bx−n, then \[x^2 \frac{d^2 y}{d x^2} =\]
If y2 = ax2 + bx + c, then \[y^3 \frac{d^2 y}{d x^2}\] is
