Basic formulas:
- (e^x)' = e^x (the most important property: it returns to itself)
- (a^x)' = a^x · log a (natural logarithm log = ln)
- (log x)' = 1/x (x > 0)
- (log_a x)' = 1/(x log a)
Example ① Differentiate the following.
(1) y = e^{3x} (2) y = log(x²+1) (3) y = 2^x
Solution:
(1) e^{3x}·3 = 3e^{3x}
(2) (1/(x²+1))·2x = 2x/(x²+1)
(3) 2^x · log 2
(1) y = e^{3x} (2) y = log(x²+1) (3) y = 2^x
Solution:
(1) e^{3x}·3 = 3e^{3x}
(2) (1/(x²+1))·2x = 2x/(x²+1)
(3) 2^x · log 2
Key Point
- e^x remains e^x when differentiated (this explains the special constant e)
- When the expression inside log is complicated, logarithmic differentiation (next lesson) is useful
Practice Problems
- Find y' for y = e^{x²}
- Find y' for y = log(2x+5)
Answers and Explanations
- Answer:2x·e^{x²}
- Answer:2/(2x+5)