If p(x)= ln((x - a)(x - b)(x - c))and a, b, c are constants, then $p ^ { \prime } ( x ) = \frac { 1 } { x - a } + \frac { 1 } { x - b } + \frac { 1 } { x - c }$ .

Question

Quizplus · Accepted Answer

The derivative of $p(x) = \ln((x - a)(x - b)(x - c))$ can be found using the chain rule and the property of logarithms that allows the logarithm of a product to be written as the sum of the logarithms. Differentiating $\ln(x - a)$, $\ln(x - b)$, and $\ln(x - c)$ individually with respect to $x$ gives $\frac{1}{x - a}$, $\frac{1}{x - b}$, and $\frac{1}{x - c}$ respectively, leading to the given expression for $p'(x)$.

If P(x)= Ln((x - A)(x - B)(x - C))and A p′(x)=1x−a+1x−b+1x−cp ^ { \prime } ( x ) = \frac { 1 } { x - a } + \frac { 1 } { x - b } + \frac { 1 } { x - c }p′(x)=x−a1​+x−b1​+x−c1​

If P(x)= Ln((x - A)(x - B)(x - C))and A $p ^ { \prime } ( x ) = \frac { 1 } { x - a } + \frac { 1 } { x - b } + \frac { 1 } { x - c }$