Gradient of $y=\ln x$

Gradient of y=\ln x

\frac{dy}{dx} \ln x = \frac{1}{x}

Gradient of y=\log{ax}

• y=\log a + \log x
• \log a is a constant, so its derivative is 0.
• Find the derivative:
  • y=\log x
  • \frac{dy}{dx} \log x = \frac{1}{x}

So the gradient of y=\log{ax} is \frac{1}{x} as well.

Gradient of y=k\ln x

• Find the derivative:
  • y=k\ln x
  • \frac{dy}{dx} k\ln x = k \frac{1}{x} = \frac{k}{x}

So the gradient of y=k\ln x is \frac{k}{x}.