Question

regard y as the independent variable and x as the dependent variable and use implicit differentiation to find $\frac{dx}{dy}$. y sec(x) = 4x tan(y) $\frac{dx}{dy}=square$

Explanation:

Step1: Differentiate both sides

Differentiate $y\sec(x)$ and $4x\tan(y)$ with respect to $y$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$.
The derivative of $y\sec(x)$ with respect to $y$ is $\sec(x)+y\sec(x)\tan(x)\frac{dx}{dy}$ (since $\frac{d}{dy}(y) = 1$ and by chain - rule $\frac{d}{dy}(\sec(x))=\sec(x)\tan(x)\frac{dx}{dy}$).
The derivative of $4x\tan(y)$ with respect to $y$ is $4\frac{dx}{dy}\tan(y)+4x\sec^{2}(y)$ (using product - rule and $\frac{d}{dy}(\tan(y))=\sec^{2}(y)$).
So we have $\sec(x)+y\sec(x)\tan(x)\frac{dx}{dy}=4\frac{dx}{dy}\tan(y)+4x\sec^{2}(y)$.

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

Move all terms with $\frac{dx}{dy}$ to one side:
$y\sec(x)\tan(x)\frac{dx}{dy}-4\frac{dx}{dy}\tan(y)=4x\sec^{2}(y)-\sec(x)$.
Factor out $\frac{dx}{dy}$:
$\frac{dx}{dy}(y\sec(x)\tan(x) - 4\tan(y))=4x\sec^{2}(y)-\sec(x)$.

Step3: Solve for $\frac{dx}{dy}$

$\frac{dx}{dy}=\frac{4x\sec^{2}(y)-\sec(x)}{y\sec(x)\tan(x)-4\tan(y)}$

Answer:

$\frac{4x\sec^{2}(y)-\sec(x)}{y\sec(x)\tan(x)-4\tan(y)}$