If xe x y - y -sin 2 x- then find dy - dx at x -0A- undefinedB- undefinedC- undefinedD- undefined

Differentiation of the expression is: x^2∗ e^xy∗dy/dx e^xyxy e^xy dy/dx = 2 cos x ^∗ d=sin x ⇒ dy/dx (x^2 ∗ e^xy 1) =2cos x^∗ sin x e^xy xye^xy ⇒ dy/dx ...

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Byju's Answer

Standard IX

Mathematics

Absolute Value

If xe x y - y...

Question

If $x e^{x} y - y = s i n^{2} x$ , then find $\frac{d y}{d x}$ at x = 0

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Solution

Differentiation of the expression is:
$x^{2} * e^{x y} * \frac{d y}{d x} e^{x y} x y e^{x y} - \frac{d y}{d x} = 2 c o s x^{} d = s i n x \Rightarrow \frac{d y}{d x} (x^{2} e^{x y} - 1) = 2 c o s x^{} s i n x - e^{x y} x y e^{x y} \Rightarrow \frac{d y}{d x} = \frac{2 c o s x^{} s i n x - e^{x y} x y e^{x y}}{x^{2} * e^{x y} - 1}$
Now, by substituting x = 0 into the expression above,
$\Rightarrow \frac{d y}{d x} = \frac{(0 - 10)}{(0 - 1)} T h u s, \frac{d y}{d x} = 1$

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If xexy−y=sin2x, then find dydx at x = 0

Differentiation of the expression is: x2∗exy∗dydxexyxyexy−dydx=2cosx∗d=sinx⇒ dydx(x2∗exy−1)=2cosx∗sinx−exyxyexy⇒ dydx=2 cos x∗sin x−exy xyexyx2∗exy−1 Now, by substituting x = 0 into the expression above, ⇒dydx=(0−1 0)(0−1)Thus, dydx=1

If $x e^{x} y - y = s i n^{2} x$ , then find $\frac{d y}{d x}$ at x = 0