Find the equation of the line tangent to the curve $f(x)=\ln x^{2}$ at the point $(1,0)$.
A) $y=2 x-2$
B) $y=x-1$
C) $y=\frac{1}{x}(x-1)$
D) $y=\frac{2}{x}(x-1)$

Question

Quizplus · Accepted Answer

The derivative of $f(x)=\ln x^{2}$ is $f'(x)=\frac{2x}{x^{2}}=\frac{2}{x}$. So the slope of the tangent line at $(1,0)$ is $f'(1)=\frac{2}{1}=2$. The equation of the tangent line is therefore $y-0=2(x-1)$, or $y=2x-2$.

Find the Equation of the Line Tangent to the Curve f(x)=ln⁡x2f(x)=\ln x^{2}f(x)=lnx2

Find the Equation of the Line Tangent to the Curve $f(x)=\ln x^{2}$