given the equation 5sin²x + 4cos²y = tan(4x² + y²), find dy/dx by implicit differentiation.

Explanation:

Step1: Differentiate both sides

Differentiate $5\sin^{2}x + 4\cos^{2}y$ and $\tan(4x^{2}+y^{2})$ with respect to $x$. For the left - hand side: The derivative of $5\sin^{2}x$ using the chain rule: Let $u = \sin x$, then $y = 5u^{2}$. $\frac{dy}{du}=10u$ and $\frac{du}{dx}=\cos x$, so $\frac{d}{dx}(5\sin^{2}x)=10\sin x\cos x$. The derivative of $4\cos^{2}y$ with respect to $x$ using the chain rule: Let $v=\cos y$, then $z = 4v^{2}$. $\frac{dz}{dv}=8v$ and $\frac{dv}{dx}=-\sin y\frac{dy}{dx}$, so $\frac{d}{dx}(4\cos^{2}y)=8\cos y(-\sin y)\frac{dy}{dx}=- 8\cos y\sin y\frac{dy}{dx}$. For the right - hand side: Let $w = 4x^{2}+y^{2}$, then $z=\tan w$. $\frac{dz}{dw}=\sec^{2}w$ and $\frac{dw}{dx}=8x + 2y\frac{dy}{dx}$. So $\frac{d}{dx}(\tan(4x^{2}+y^{2}))=\sec^{2}(4x^{2}+y^{2})(8x + 2y\frac{dy}{dx})$. So we have $10\sin x\cos x-8\cos y\sin y\frac{dy}{dx}=\sec^{2}(4x^{2}+y^{2})(8x + 2y\frac{dy}{dx})$.

Step2: Isolate $\frac{dy}{dx}$

Expand the right - hand side: $10\sin x\cos x-8\cos y\sin y\frac{dy}{dx}=8x\sec^{2}(4x^{2}+y^{2})+2y\sec^{2}(4x^{2}+y^{2})\frac{dy}{dx}$. Move the terms with $\frac{dy}{dx}$ to one side: $-8\cos y\sin y\frac{dy}{dx}-2y\sec^{2}(4x^{2}+y^{2})\frac{dy}{dx}=8x\sec^{2}(4x^{2}+y^{2}) - 10\sin x\cos x$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(-8\cos y\sin y - 2y\sec^{2}(4x^{2}+y^{2}))=8x\sec^{2}(4x^{2}+y^{2}) - 10\sin x\cos x$. Then $\frac{dy}{dx}=\frac{10\sin x\cos x - 8x\sec^{2}(4x^{2}+y^{2})}{8\cos y\sin y+2y\sec^{2}(4x^{2}+y^{2})}$.

Answer:

$\frac{10\sin x\cos x - 8x\sec^{2}(4x^{2}+y^{2})}{8\cos y\sin y+2y\sec^{2}(4x^{2}+y^{2})}$