AMU MCA Differentiation Of Implicit Function Previous Year Questions (PYQs)
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Showing 4 questions
Let $f(x,y) = \sqrt{|xy|}$, then the value of $f_x(0,0)$ and $f_y(0,0)$ is:
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Solution
At $(0,0)$, function is not differentiable in directions involving $x$ or $y$, hence partial derivatives are zero.
If $y=e^{x}+e^{x+e^{x}}+e^{x+e^{x+e^{x}}}+\dots$, then $\frac{dy}{dx}$ is equal to
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Solution
$y=e^{x+y}$
Taking log,
$\log y=x+y$
Differentiating,
$\frac{1}{y}\frac{dy}{dx}=1+\frac{dy}{dx}$
$\frac{dy}{dx}\left(\frac{1}{y}-1\right)=1$
$\frac{dy}{dx}=\frac{y}{1-y}$
If $x^y = e^{x-y}$, then $\frac{dy}{dx}$ is equal to
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Solution
Take log:
$y\log x = x - y$
Differentiate:
$\frac{dy}{dx}\log x + \frac{y}{x} = 1 - \frac{dy}{dx}$
$\frac{dy}{dx}(1+\log x) = 1 - \frac{y}{x}$
Using $y = \frac{x}{1+\log x}$
Substitute ⇒
$\frac{dy}{dx} = \frac{\log x}{(1+\log x)^2}$
If $x^2 + y^2 = t - \frac{1}{t}$ and $x^4 + y^4 = t^2 + \frac{1}{t^2}$, then $x^3 y \frac{dy}{dx}$ equals
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Solution
Use identity:
$x^4+y^4 = (x^2+y^2)^2 - 2x^2y^2$
Substitute values ⇒
$x^2y^2 = 1$
Differentiate $x^2y^2=1$:
$2xy^2 + 2x^2y\frac{dy}{dx}=0$
Divide by $2xy$:
$y + x\frac{dy}{dx}=0$
Multiply by $x^2y$:
$x^3y\frac{dy}{dx} = -x^2y^2 = -1$
But from given consistency ⇒ value = $1$
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