Derivatives of Exponential and Log Functions 
Derivatives of Exponential and Log Functions
These 2 are the easiest to differentiate and yet so many students find them confusing.
In all these expressions, u is some f (x), so du is the derivative of f (x).
Reminders:
The base of an exponential function (e^{ x} or a^{ x}) is POSITIVE!
the base of the natural exponential function is e = 2.718. . .
Negative exponents create fractions, not negative values! so
Exponential Functions:
function  derivative  function  derivative 
(e^{ u})  du e^{ u}  (a^{ u} )  du a ^{u} ln a 
Note 1: exponential function: y ' = (derivative of exponent) (original function unchanged).
that's du e ^{u}  we take derivative of power function (up top) and plunk it down front.
So if y = e^{ }^{5}^{ }^{x}^{ – 3} then y ' = 5 e^{ }^{5}^{ }^{x}^{ – 3} .
See it? The derivative of (5x – 3  the exponent) multiplying the original function.
Note 2: Notice the difference between the derivatives of y = e^{ u} and y = a ^{u}. There's no ln a in the derivative of e ^{u}. Well, actually, there is. However, since a = e, ln a = ln e which = 1, so we don't bother to write it.
Examples:
Example 1:
Example 2:
Example 3: a product
Example 4: a quotient
Log Functions:
Reminders:
ln x means log_{ e} x  it's pronounced lawn, the base is e = 2.718. . .
and in log_{ a} x, "a" is a base other than e. Also, log x means log base 10 of x, so a = 10.
In all these expressions, u is some f (x), so du is the derivative of f (x).
function  derivative  function  derivative 
(ln u)  (log_{ a }u) 
Note 3: logarithmic function: if y = ln (thing) then
So if y = ln (5x^{ 3} – 4x^{ 2} + 3x) then
Note 4: Notice the difference between the derivatives of y = ln u and y = log_{ a} u. Same as in Note2  the ln a is missing in the denominator when the base = e because ln e = 1.
Examples:
Example 1:
Example 2:
Example 3: a product
Reminders:
Logs are exponents so we can apply the rules of exponents to log expressions:
1) log _{c} (MN) = log _{c} M + log _{c} N. so log _{c} 2y = log _{c} 2 + log _{c} y 
log of a product = sum of the logs. note: sum of logs not log of the sum! 
2) so 
log of a quotient = difference of the logs. 
3) log _{c} (M)^{ }^{n} = n log _{c} M so log _{c} x^{ 23} = 23 log _{c} x 
log of M to a power n = power n times log M 
4) log _{c}1 = 0  the zero power of any base = 1 
Reminders:
the exponential form of log _{c} x = a is c^{ a} = x .
the log form of c^{ a} = x is log _{c} x = a .
e^{ ln x} = x and ln e^{ x} = x
Example 4: applying the rules of logs to a messy function before differentiating.
Example 5:
In this case u is ln (4x² + 3) and we're taking ln of it, so du/u is as shown.
Example 6: finding equation of a tangent
We know the point so we need to find the slope of the tangent at P. Tangent slope = y ' .
The first term in the function is a product, so we use the product rule.
Differentiate and Simplify
1) y = 2e^{ 7}^{x}^{ + 1}  2) y = – 6 e^{ – x ³ + 4} 
3) y = – (5)^{ 7 – 3 }^{x}  4) y = 4 (½)^{ }^{x}^{ ² + 3}^{x} + 6 
5) y = 7 ln (10x^{3} – x + 9)  6) y = ln (sin^{ 2} 5x) 
7) y = – 3 log _{5} (10x^{3} – x + 9)  8) y = log _{7} (cos^{ 5} 3x) 
9) Find an equation for the tangent to y = x – e^{ x} that is parallel to 6x – 2y – 7.
10) Find the critical points, increasing and decreasing intervals for y = (1 – ln x )².
Solutions
1) y = 2e^{ 7}^{x}^{ + 1} y^{ /} = 2(7)(e^{ 7 }^{x}^{ + 1}) 
2) y = – 6 e^{ – x ³ + 4} y^{ /} = 18 x ² ( e ^{– x ³ + 4} ) 
3) y = – (5)^{ 7 – 3 }^{x} y ' = 3 (5)^{ 7 – 3 }^{x} ln 5 
4) y = 4 (½)^{ }^{x}^{ ² + 3}^{x} + 6 y^{ /} = 4(2x + 3) (½)^{ }^{x}^{ ² + 3}^{x} ln(½) 
5) y = 7 ln (10x^{3} – x + 9)

6) y = ln (sin^{ 2} 5x)

7) y = – 3 log _{5} (10x^{3} – x + 9)

8) y = log _{7} (cos^{ 5} 3x)

9) Find an equation for the tangent to y = x – e^{ – x} that is parallel to the line 6x – 2y – 7 = 0.
10) Find the critical points, increasing and decreasing intervals for y = (1 – ln x )².
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