Question

differentiate the function.
y = \frac{1}{(7x - 9)^2}
\frac{dy}{dx} = \square

Explanation:

Step1: Rewrite the function

Rewrite \( y = \frac{1}{(7x - 9)^2} \) as \( y=(7x - 9)^{-2} \) using the negative exponent rule \( \frac{1}{a^n}=a^{-n} \).

Step2: Apply the chain rule

The chain rule states that if \( y = u^n \) and \( u = f(x) \), then \( \frac{dy}{dx}=n\cdot u^{n - 1}\cdot\frac{du}{dx} \). Here, \( u = 7x-9 \), \( n=-2 \). First, find the derivative of \( u \) with respect to \( x \): \( \frac{du}{dx}=7 \). Then, find the derivative of \( y \) with respect to \( u \): \( \frac{dy}{du}=-2u^{-3} \).

Step3: Multiply the derivatives

Using the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \), substitute the values: \( \frac{dy}{dx}=-2u^{-3}\cdot7 \). Substitute back \( u = 7x - 9 \): \( \frac{dy}{dx}=-14(7x - 9)^{-3} \). Rewrite using positive exponents: \( \frac{dy}{dx}=-\frac{14}{(7x - 9)^3} \).

Answer:

\( -\dfrac{14}{(7x - 9)^3} \)