Chapter 3 | Derivatives
327
dy dx =
(3.34)
b x ln b .
g ( x ) ,
More generally, if h ( x ) = b
then
(3.35)
h ′( x ) = b g ( x ) g ′( x )ln b .
Proof If y = log b x , then b
y = x . It follows that ln( b y ) = ln x . Thus y ln b = ln x . Solving for y , we have y = ln x ln b .
Differentiating and keeping in mind that ln b is a constant, we see that dy dx = 1 x ln b . The derivative in Equation 3.33 now follows from the chain rule. If y = b x , then ln y = x ln b . Using implicit differentiation, again keeping in mind that ln b is constant, it follows that 1 y dy dx = ln b . Solving for dy dx
and substituting y = b x , we see that dy dx = y ln b = b x ln b . The more general derivative ( Equation 3.35 ) follows from the chain rule. □ Example 3.79 Applying Derivative Formulas
Find the derivative of h ( x ) = 3 x 3 x +2 .
Solution Use the quotient rule and Derivatives of General Exponential and Logarithmic Functions . h ′( x ) = 3 x ln3(3 x +2)−3 x ln3(3 x ) (3 x +2) 2 Apply the quotient rule. = 2·3 x ln3 (3 x +2) 2 Simplify.
Example 3.80 Finding the Slope of a Tangent Line
Find the slope of the line tangent to the graph of y = log 2 (3 x +1) at x =1.
Solution
Made with FlippingBook - professional solution for displaying marketing and sales documents online