1. Differentiating exponential functions.
Predictions
Exponential growth,
Tangent slopes are positive, and like the function itself, the tangent slopes increase in value to the right.
𝑦= 0.5 𝑥
Exponential decay,
Tangent slopes are negative and like the function itself, the tangent slopes approach zero as we move to the right.
𝑦= 2 𝑥
Is the derivative of the exponential function also an exponential function??????

2. Can the power rule and the chain rule be used to differentiate an exponential function?
No!
𝐼𝑓 𝑦= 1 2 𝑥
Exponential growth
𝐷𝑜𝑒𝑠 𝑑𝑦 𝑑𝑥 =(1)(𝑥) 2 𝑥−1 𝑧𝑒𝑟𝑜 ??????
No!
We need a rule for differentiating exponential functions.
Chain rule step
Exponential functions do not have any tangents with a zero slope value. None of their tangents are horizontal. Differentiating with the power rule and the chain rule suggests that all the tangents are horizontal.

3. 𝑓 𝑥 =4 𝑥 3 𝑎𝑛𝑑 𝑑𝑓(𝑥) 𝑑𝑥 =12 𝑥 2
are polynomial functions.
𝑔 𝑥 = 3𝑥 2𝑥+5 𝑎𝑛𝑑 𝑑𝑔(𝑥) 𝑑𝑥 = 3 2𝑥+5 −(2)(3𝑥) 2𝑥+5 2
are rational functions
𝑡 𝑥 =𝑠𝑖𝑛𝑥 𝑎𝑛𝑑 𝑑𝑡(𝑥) 𝑑𝑥 =𝑐𝑜𝑠𝑥
are trigonometric functions
Then should we expect the derivatives of an exponential equation to be exponential with the same base?
Yes!
For b>0, if 𝑦=𝑎 𝑏 𝑢 , 𝑑𝑦 𝑑𝑥 = 𝑎 𝑏 𝑢 ln⁡(𝑏) 𝑑𝑢 𝑑𝑥

4. Examples: 3. ℎ 𝑥 =𝑒 𝑥 1. 𝑓 𝑥 =3 𝑥 2. 𝑔 𝑥 =5 𝑥 𝑑𝑓(𝑥) 𝑑𝑥 = 3 𝑥 ln3
1. 𝑓 𝑥 =3 𝑥
3. ℎ 𝑥 =𝑒 𝑥
2. 𝑔 𝑥 =5 𝑥
𝑑𝑓(𝑥) 𝑑𝑥 = 3 𝑥 ln3
𝑑𝑔(𝑥) 𝑑𝑥 = 5 𝑥 ln5
𝑑ℎ(𝑥) 𝑑𝑥 = 𝑒 𝑥 ln(e) or 𝑒 𝑥 [one]
𝑑ℎ(𝑥) 𝑑𝑥 = 𝑒 𝑥
In summary, for exponential functions; b>0, b≠1
𝑦=𝑎 𝑒 𝑥
𝑑𝑦 𝑑𝑥 =𝑎 𝑒 𝑥
𝑦=𝑎 𝑏 𝑥
𝑑𝑦 𝑑𝑥 =𝑎 𝑏 𝑥 ln(b)

5. Recall, that the exponent is the variable for exponential functions.
If another function has been substituted in for the exponent “x”, then the exponential function is a composite function and the chain rule step must then be used when differentiating.
Given:
𝑦= 𝑥 ,
we will determine the derivative two different ways
Method one:
Method two:
𝑑𝑦 𝑑𝑥 = 𝑥 [ln0.5]
𝑑𝑦 𝑑𝑥 = 2 −1 𝑥 [ln 2 −1 ]
𝑑𝑦 𝑑𝑥 = 2 −𝑥 [-1][ln2]
𝑦= 𝑥 or
𝑦= 2 −𝑥
Chain rule step
The same formula
𝑑𝑦 𝑑𝑥 = 2 −𝑥 [ln2][-1]

6. More examples: 𝑦=−3 5 2𝑥 𝑦=−3 𝑒 2𝑥 𝑑𝑦 𝑑𝑥 =−3 5 2𝑥 [𝑙𝑛5][2]1.
𝑦=− 𝑥 2.
𝑦=−3 𝑒 2𝑥
Constant multiple rule
Chain rule step
Constant
multiple rule
Chain rule step
𝑑𝑦 𝑑𝑥 =− 𝑥 [𝑙𝑛5][2]
𝑑𝑦 𝑑𝑥 =−3 𝑒 2𝑥 [𝑙𝑛 𝑒 ][2]
Recall ln e = 1
Rule for differentiating exponential functions
Rule for differentiating exponential functions
𝑑𝑦 𝑑𝑥 =−3 𝑒 2𝑥 [2] or -6 𝑒 2𝑥