Question

find the derivative of the following function.

$y = \ln(5x^{6} - 7x^{5})$

Explanation:

Step1: Apply the chain rule

Let \(u = 5x^{6}-7x^{5}\), then \(y=\ln(u)\). The chain rule states that \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
First, find \(\frac{dy}{du}\):
Since \(y = \ln(u)\), \(\frac{dy}{du}=\frac{1}{u}\).

Step2: Find \(\frac{du}{dx}\)

Since \(u = 5x^{6}-7x^{5}\), using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(\frac{du}{dx}=5\times6x^{5}-7\times5x^{4}=30x^{5}-35x^{4}\).

Step3: Substitute back

Substitute \(u = 5x^{6}-7x^{5}\), \(\frac{dy}{du}=\frac{1}{u}\) and \(\frac{du}{dx}=30x^{5}-35x^{4}\) into \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\):
\(\frac{dy}{dx}=\frac{30x^{5}-35x^{4}}{5x^{6}-7x^{5}}\).
Factor out \(5x^{4}\) from the numerator and \(x^{5}\) from the denominator:
\(\frac{dy}{dx}=\frac{5x^{4}(6x - 7)}{x^{5}(5x - 7)}=\frac{5(6x - 7)}{x(5x - 7)}=\frac{30x-35}{5x^{2}-7x}\).

Answer:

\(\frac{30x - 35}{5x^{2}-7x}\)