Give an example of each type of differentiation we did in Ch. 5. Use each formula and make sure your example problem illustrates how the formula is used.
Expanding and Condensing natural logs are easy, when it's multiplication you add the terms, then it's division you subtract the terms to expand and you work it backwards when condensing...this isn't exactly differentiating anything but it makes things easier when you are differentiating.
In Chapter 5, we learned many different types of differentiation.
ReplyDeleteSection 5.1
Examples:
Expand.
1) ln 1/e = ln1 -lne
2) ln z(z-1)^2
= lnz + ln(z-1)^2
= lnz + 2ln(z-1)
Examples:
1) ln(x-2) - ln(x+2)
= ln (x-2)/(x+2)
2) 3/2 [ln(x^2+1) - ln(x+1) - ln(x-1)]
= 3/2 ln (x^2+1)/(x+1)(x-1)
Section 5.2
Formula: d/dx lnu = 1/u X du
Examples:
1) d/dx ln(x^2+1) = 1/x^2+1 X 2x
= 2x/x^2+1
2) d/dx (lnx)^3 = 3lnx)^2 X 1/x X 1
= 3(lnx)^2/x
3) d/dx lnx^4 = 1/x^4 X 4x^3
= 4/x
Section 5.4
Formulas: e^lnx = x
lne^x = x
Examples:
1) 7 = e^x+1
ln7 = lne^x+1
ln7 = x+1
x = ln7-1
2) ln(3x-2) = 6
e^6 = 3x-2
e^6 + 2 = 3x
x = e^6+2/3
Next Sections:
Formulas: d/du e^u = e^u X du
d/du #^u = #^u ln# X du
Examples:
1) d/dx e^3x^4 = e^3x^4 X 12x^3
= 12x^3e^3x^4
2) d/dx 5^x = 5^xln5 X 1
= 5^xln5
More Examples:
1) y = log 4 (5x+1)
= 1/(5x+1)ln4 X 5
= 5/(5x+1)ln4
**The problem is log base 4...
In chapter five we learned a few different
ReplyDeletetypes of differentiations.
one is the derivative of a
natural log.
formula: d/dx lnu=1/u x du
example:
d/dx ln2x=1/2x x 2
the twos cancel and you are left with
1/x as your final answer.
another type that we learned is
e differentiation.
formula: d/dx e^u= e^u X du
example: d/dx e^x^2= e^x^2 X 2x= 2xe^x^2
one more type is log
differentiation.
formula: d/dx log a^u= 1/ulna X du
example:
log 5 x^2= 1/ x^2ln5 X 2x= 2/xln5
that is all!
In Chapter 5, we learned many different types of differentiation.
ReplyDeleteWith logarithms
Examples:
Expand.
1) ln 1/e = ln1 -lne
2) ln z(z-1)^2
= lnz + ln(z-1)^2
= lnz + 2ln(z-1)
Examples:
1) ln(x-2) - ln(x+2)
= ln (x-2)/(x+2)
2) 3/2 [ln(x^2+1) - ln(x+1) - ln(x-1)]
= 3/2 ln (x^2+1)/(x+1)(x-1)
Formula: d/dx lnu = 1/u X du
Examples:
1) d/dx ln(x^2+1) = 1/x^2+1 X 2x
= 2x/x^2+1
2) d/dx (lnx)^3 = 3lnx)^2 X 1/x X 1
= 3(lnx)^2/x
3) d/dx lnx^4 = 1/x^4 X 4x^3
= 4/x
Formulas: e^lnx = x
lne^x = x
Examples:
1) 7 = e^x+1
ln7 = lne^x+1
ln7 = x+1
x = ln7-1
2) ln(3x-2) = 6
e^6 = 3x-2
e^6 + 2 = 3x
x = e^6+2/3
Formulas: d/du e^u = e^u X du
d/du #^u = #^u ln# X du
Examples:
1) d/dx e^3x^4 = e^3x^4 X 12x^3
= 12x^3e^3x^4
2) d/dx 5^x = 5^xln5 X 1
= 5^xln5
1) y = log 4 (5x+1)
= 1/(5x+1)ln4 X 5
= 5/(5x+1)ln4
In chapter five we learned a few different
ReplyDeletetypes of differentiations.
one is the derivative of a
natural log.
formula: d/dx lnu=1/u x du
another type that we learned is
e differentiation.
formula: d/dx e^u= e^u X du
one more type is log
differentiation.
formula: d/dx log a^u= 1/ulna X du
Different types of differentiation from Ch. 5:
ReplyDeleteExpanding and Condensing natural logs are easy, when it's multiplication you add the terms, then it's division you subtract the terms to expand and you work it backwards when condensing...this isn't exactly differentiating anything but it makes things easier when you are differentiating.
*d/dx ln u=1/u * du
Example :
ln ( 3x+2 )
( 1 ) / ( 3x+2 ) * 3
( 3 ) / ( 3x+2 )
*d/dx e^u= e^u * du
Example:
e^4x^2
e^4x^2 * 8x
8x e^4x^2
*#^u=#^uln# * du
Example:
3^x2
3^x2ln3 * 2x
*d/dx log a^u = 1/ulna * du
Example:
y = log 4 ( 5x+1 )
1 /( 5x+1 )ln4 X 5
5 /( 5x+1 )ln4