Question

if (y = ln(3x + 4y)), then (\frac{dy}{dx}=) a) (\frac{1}{3x + 4y}) b) (\frac{3}{3x + 4y}) c) (\frac{7}{3x + 4y}) d) (\frac{4}{3x + 4y})

Explanation:

Step1: Differentiate using chain - rule

If $y = \ln(u)$, then $\frac{dy}{dx}=\frac{1}{u}\cdot\frac{du}{dx}$. Here $u = 3x + 4y$. So $\frac{dy}{dx}=\frac{1}{3x + 4y}\cdot(3 + 4\frac{dy}{dx})$.

Step2: Expand the right - hand side

$\frac{dy}{dx}=\frac{3}{3x + 4y}+\frac{4}{3x + 4y}\cdot\frac{dy}{dx}$.

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

$\frac{dy}{dx}-\frac{4}{3x + 4y}\cdot\frac{dy}{dx}=\frac{3}{3x + 4y}$.

Step4: Factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(1 - \frac{4}{3x + 4y})=\frac{3}{3x + 4y}$.

Step5: Simplify the left - hand side

$\frac{dy}{dx}(\frac{3x + 4y-4}{3x + 4y})=\frac{3}{3x + 4y}$.

Step6: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{3}{3x + 4y - 4}$.

Answer:

$\frac{3}{3x + 4y-4}$