Question

$\begin{aligned} &\text { A) } x e^{x y} \mp y+\sin ^{2} x\\\ &\text { At } x=0, \quad y=0\\\ &\text { Differentiating wrt } \mathrm{x} \text {, }\\\ &\mathrm{e}^{\mathrm{xy}}+\mathrm{xe}^{\mathrm{xy}}\left(\mathrm{y}+\mathrm{xy}^{\prime}\right)=\mathrm{y}^{\prime}+\sin 2 \mathrm{x} \end{aligned}$ At \(x=0\) \(\quad \Rightarrow 1=y^{\prime}\) B) Conceptual C) Conceptual D) $\begin{aligned} y &=\sin ^{-1} \cos \sin ^{-1} x+\cos ^{-1} \sin \cos ^{-1} x \\ &=\sin ^{-1} \sqrt{1-x^{2}}+\cos ^{-1} \sqrt{1-x^{2}} \\\ &=\frac{\pi}{2} \end{aligned}$

Short answer

Based on the step-by-step solution provided, create a short answer below. Given the function \(A) x e^{xy} \mp y+\sin^2 x\) and the point \((x, y) = (0, 0)\), we used the chain rule, product rule, and trigonometric formulae to differentiate the function with respect to x. After finding the first derivative, we plugged in the given point to evaluate it, obtaining the value of the first derivative at \((0, 0)\) as \(1\).

Step-by-step solution

Understand the given function and point

We are given the function \(A) x e^{xy} \mp y+\sin^2 x\) and a point \((x, y) = (0, 0)\). Our goal is to find the first derivative \(\frac{dy}{dx}\) at this point.

Differentiate the function with respect to x

Using the chain rule and product rule, we differentiate the function with respect to x as follows: \(\frac{dy}{dx}=\frac{d}{dx}(xe^{xy} \mp y+\sin^2 x)\) Using the product rule for the first term: \(\frac{d}{dx}(xe^{xy}) = e^{xy} + x(\frac{d}{dx}(e^{xy}))\) The chain rule is then applied to find \(\frac{d}{dx}(e^{xy})\): \(\frac{d}{dx}(e^{xy}) = e^{xy}(\frac{d}{dx}(xy)) = e^{xy}(y + x\frac{dy}{dx})\) Now, differentiate the rest of the function terms: \(\frac{d}{dx}(\mp y) = \mp \frac{dy}{dx}\) \(\frac{d}{dx}(\sin^2 x) = 2\sin x \cos x\) Therefore, combining all terms, we get: \(\frac{dy}{dx} = e^{xy} + x(y + x\frac{dy}{dx})e^{xy} \mp \frac{dy}{dx} + 2\sin x \cos x\)

Evaluate the first derivative at the given point

Now, we plug in the point \((x, y) = (0, 0)\) into the derivative: \(\frac{dy}{dx}(0,0) = e^{(0)(0)} + (0)(0 + (0)\frac{dy}{dx}(0,0))e^{(0)(0)} \mp \frac{dy}{dx}(0,0) + 2\sin(0) \cos (0)\) Simplify the terms: \(\frac{dy}{dx}(0,0) = 1 \mp \frac{dy}{dx}(0,0)\) Now, solve for \(\frac{dy}{dx}(0,0)\): \(\frac{dy}{dx}(0,0) = 1\) So, the value of the first derivative at the point \((0, 0)\) is \(1\).