Calculus Volume 1

Chapter 3 | Derivatives

323

b x + h − b x h b x b h − b x h b x ( b h −1) h

B ′( x ) = lim h →0

Apply the limit definition of the derivative.

b x + h = b x b h .

= lim

Note that

h →0

b x .

= lim

Factor out

h →0

b h −1 h

= b x lim h →0

Apply a property of limits.

b 0+ h − b 0 h

b h −1

= b x B ′(0)

Use B ′(0) = lim h →0 h . We see that on the basis of the assumption that B ( x ) = b x is differentiable at 0, B ( x ) is not only differentiable everywhere, but its derivative is (3.29) B ′( x ) = b x B ′(0). For E ( x ) = e x , E ′(0) =1. Thus, we have E ′( x ) = e x . (The value of B ′(0) for an arbitrary function of the form B ( x ) = b x , b >0, will be derived later.) = lim h →0

Theorem 3.14: Derivative of the Natural Exponential Function Let E ( x ) = e x be the natural exponential function. Then E ′( x ) = e x . In general, d dx ⎛ ⎝ e g ( x ) ⎞ ⎠ = e g ( x ) g ′( x ).

Example 3.74 Derivative of an Exponential Function Find the derivative of f ( x ) = e tan(2 x ) .

Solution Using the derivative formula and the chain rule,

f ′( x ) = e tan(2 x ) d dx ⎛

⎝ tan(2 x ) ⎞ ⎠ = e tan(2 x ) sec 2 (2 x ) ·2.

Example 3.75 Combining Differentiation Rules

Find the derivative of y = e x 2 x .

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