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27 Questions around this concept.
For the curve $y=3 \sin \theta \cos \theta, \mathrm{x}=e^\theta \sin \theta, 0 \leq \theta \leq \pi$, the tangent is parallel to $x$-axis when $\theta$ is :
If $\mathrm{x^{y}+y^{x}= 2,\, then\, \frac{dy}{dx}\, at\; x= 1\, equals}$
Differentiation of Implicit Function
If variables $x$ and $y$ are connected by a relation of the form $f(x, y)=0$ and it is not possible or convenient to express y as a function of x i.e., in the form $\mathrm{y}=\Phi(\mathrm{x})$, then y is said to be an implicit function.
To find $\frac{d y}{d x}$ in such a case, we differentiate both sides of the given relation concerning x keeping in mind that the derivative of $\Phi(\mathrm{y})$ concerning x is $\frac{d \phi}{d y} \times \frac{d y}{d x}$.
For example
$
\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}, \frac{d}{d x}\left(y^2\right)=2 y \frac{d y}{d x}
$
It should be noted that $\frac{d}{d y}(\sin y)=\cos y$ but $\frac{d}{d x}(\sin y)=\cos y \frac{d y}{d x}$
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