Question

If $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x+3}{3-2x}\right)$, then $\frac{dy}{dx}$ at $x=1$ is

• A. $6 \sin(\log 5)$
• B. $5 \sin(\log 6)$
• C. $12 \sin(\log 5)$
• D. $5 \sin(\log 12)$

Hint:

Remember the chain rule: if $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$. Also, be proficient with the quotient rule for differentiation.

Answer 1

The Correct Option is A

Step 1: Identify the given functions and the expression to differentiate. We are given $f'(x) = \sin(\log x)$ and $y = f\left(\frac{2x+3}{3-2x}\right)$. We need to find $\frac{dy}{dx}$ at $x=1$. Let $u = \frac{2x+3}{3-2x}$. Then $y = f(u)$.
Step 2: Apply the chain rule to find $\frac{dy}{dx}$. \[ \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = f'(u) \cdot \frac{du}{dx} \]
Step 3: Calculate $f'(u)$. Since $f'(x) = \sin(\log x)$, replacing $x$ with $u$ gives: \[ f'(u) = \sin(\log u) = \sin\left(\log\left(\frac{2x+3}{3-2x}\right)\right) \]
Step 4: Calculate $\frac{du}{dx}$ using the quotient rule for $u = \frac{2x+3}{3-2x}$. The quotient rule is $\left(\frac{p}{q}\right)' = \frac{p'q - pq'}{q^2}$. Here, $p = 2x+3 \implies p' = 2$. And $q = 3-2x \implies q' = -2$. \[ \frac{du}{dx} = \frac{(2)(3-2x) - (2x+3)(-2)}{(3-2x)^2} \] \[ \frac{du}{dx} = \frac{6-4x - (-4x-6)}{(3-2x)^2} \] \[ \frac{du}{dx} = \frac{6-4x + 4x + 6}{(3-2x)^2} \] \[ \frac{du}{dx} = \frac{12}{(3-2x)^2} \]
Step 5: Substitute $f'(u)$ and $\frac{du}{dx}$ back into the chain rule formula. \[ \frac{dy}{dx} = \sin\left(\log\left(\frac{2x+3}{3-2x}\right)\right) \cdot \frac{12}{(3-2x)^2} \]
Step 6: Evaluate $\frac{dy}{dx}$ at $x=1$. First, find $u$ at $x=1$: \[ u(1) = \frac{2(1)+3}{3-2(1)} = \frac{2+3}{3-2} = \frac{5}{1} = 5 \] Now substitute $x=1$ into the expression for $\frac{dy}{dx}$: \[ \left.\frac{dy}{dx}\right|_{x=1} = \sin(\log 5) \cdot \frac{12}{(3-2(1))^2} \] \[ \left.\frac{dy}{dx}\right|_{x=1} = \sin(\log 5) \cdot \frac{12}{(3-2)^2} \] \[ \left.\frac{dy}{dx}\right|_{x=1} = \sin(\log 5) \cdot \frac{12}{(1)^2} \] \[ \left.\frac{dy}{dx}\right|_{x=1} = 12 \sin(\log 5) \]