question 11 of 15, step 1 of 1 given the equation 5sin²x + 4cos²y = tan(4x² + y²), find dy/dx by implicit…

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 is $5\times2\sin x\cos x = 5\sin2x$. The derivative of $4\cos^{2}y$ using the chain - rule is $4\times2\cos y(-\sin y)\frac{dy}{dx}=-4\sin2y\frac{dy}{dx}$. For the right - hand side: Using the chain - rule, if $u = 4x^{2}+y^{2}$, then the derivative of $\tan u$ with respect to $x$ is $\sec^{2}(4x^{2}+y^{2})\left(8x + 2y\frac{dy}{dx}\right)$. So, $5\sin2x-4\sin2y\frac{dy}{dx}=\sec^{2}(4x^{2}+y^{2})\left(8x + 2y\frac{dy}{dx}\right)$.

Step2: Expand the right - hand side

Expand $\sec^{2}(4x^{2}+y^{2})\left(8x + 2y\frac{dy}{dx}\right)$ to get $8x\sec^{2}(4x^{2}+y^{2})+2y\sec^{2}(4x^{2}+y^{2})\frac{dy}{dx}$. The equation becomes $5\sin2x-4\sin2y\frac{dy}{dx}=8x\sec^{2}(4x^{2}+y^{2})+2y\sec^{2}(4x^{2}+y^{2})\frac{dy}{dx}$.

Step3: Isolate $\frac{dy}{dx}$

Move all terms with $\frac{dy}{dx}$ to one side: $-4\sin2y\frac{dy}{dx}-2y\sec^{2}(4x^{2}+y^{2})\frac{dy}{dx}=8x\sec^{2}(4x^{2}+y^{2}) - 5\sin2x$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}\left(-4\sin2y-2y\sec^{2}(4x^{2}+y^{2})\right)=8x\sec^{2}(4x^{2}+y^{2}) - 5\sin2x$. Then $\frac{dy}{dx}=\frac{8x\sec^{2}(4x^{2}+y^{2}) - 5\sin2x}{-4\sin2y-2y\sec^{2}(4x^{2}+y^{2})}=\frac{5\sin2x - 8x\sec^{2}(4x^{2}+y^{2})}{4\sin2y + 2y\sec^{2}(4x^{2}+y^{2})}$.

Answer:

$\frac{5\sin2x - 8x\sec^{2}(4x^{2}+y^{2})}{4\sin2y + 2y\sec^{2}(4x^{2}+y^{2})}$