Chapter 5: Blocks recap Flashcards
The Mean Value Theorem for Integrals
Define an inverse function
Let f be a one-to-one function with domain A and range B.
Then the inverse function f-1 has domain B and range A and is defined by:
f-1(y) = x <=> f(x) = y
for any y in B
Cancellation equations
f-1( f(x) ) = x
f( f-1(x) ) = x
Theorem on continuity of inverses
If f is a one-to-one continuous function defined on an interval, then its inverse function f-1 is also continuous.
Theorem on differentiability of inverses
d/dx (eu)
eu . du/dx
Cancellation equations for log
- loga(ax) = x
- alogax = x for every x>0
3 Log properties:
- loga(xy) = logax + logay
- loga(** x/**<strong>y</strong>) = logax - logay
- loga(xr) = r.logax
Change of base formula
logax = lnx/lna
Derivative of lnx
1/x
Derivative of ln(u) where u is a function
**1/u **. du/dx
Derivative of [ln g(x)]
g’(x)/g(x)
Derivative of ln |x|
1/x
Integral of 1/x
ln |x| + C
Integral of tanx
ln |sec x| + C
Derivative of logax
1/x.lna
Derivative of ax
ax ln a
Integral of ax
ax/lna + C a != 1
3 Steps in Logarithmic Differentiation
- Take natural logarithms of both sides of an equation y = f(x) and use the properties of logarithms to simplify.
- Differentiate implicitly with respect to x.
- Solve the resulting equation for y’
d/dx ab
0
d/dx [f(x)]b
b[f(x)]b-1 f’(x)
d/dx ag(x)
ag(x) (ln a).g’(x)
Method to solve d/dx [f(x)] g(x)
Use logarithmic differentiation