Sum and Difference Rule
The sum and difference rule is simply a continuation of the power rule. In situations where there are multiple terms in an equation all added or subtracted from one another, simply perform a differentiation on all the separate pieces. There is no need to change signs.
d [ u ± v ] d x = u ′ ± v ′ {\displaystyle {\frac {d[u\pm v]}{dx}}=u'\pm v'}
![{\displaystyle {\frac {d[u\pm v]}{dx}}=u'\pm v'}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fc36cd68414a92f501116cef0d4e4039c5e8cbf6)
Sample Question:
Find the derivative of
f
(
x
)
=
3
x
2
+
4
x
4
{\displaystyle \ f(x)=3x^{2}+4x^{4}}
f ′ ( x ) = 6 x + 16 x 3 {\displaystyle \ f'(x)=6x+16x^{3}}
Finding the Equation of a Tangent
Finding the equation of a tangent is one of the first possible application of derivatives. Below we will explain the steps to complete this kind of problem, work out a similar problem and also link a video description. There will also be practice questions.
Steps
1. First take the derivative of the given equation.
2. Use the given co-ordinates in the problem to plug into the newly derived equation in order to find the slope of the tangent line at that specific point.
3. We can now realize that a tangent line will be a straight line so we can use the y=mx + b format
4. The slope that we have discovered using the differentiated equation can be substituted from m
5. Now, we must use the given x and y co-ordinates (is y or x is not not given, use the original equation to determine their value) and substitute them into the x and y positions of our straight line equation.
6. Finally, solve for b and we have the straight line equation
Once the sloped is achieved, we may use another form to find the final equation of the tangent.
1. Set the slope equal to (y-(y value given))/(x-(x value given))
2. Solve for y by cross multiplying and other algebraic procedures.
Video
Worked out Problem
Find the equation of the tangent line of a slope that is parallel to y=3x^4+5x at (1,0).
1. First take the derivative of the given equation.
y
′
(
x
)
=
12
x
3
+
5
{\displaystyle \ y'(x)=12x^{3}+5}
2. Use the given co-ordinates in the problem to plug into the newly derived equation in order to find the slope of the tangent line at that specific point.
at
(
1
,
0
)
{\displaystyle \ (1,0)}
y
′
(
1
)
=
12
(
1
)
+
5
=
17
{\displaystyle \ y'(1)=12(1)+5=17}
3. We can now realize that a tangent line will be a straight line so we can use the y=mx + b format
y
−
y
o
=
m
(
x
−
x
o
)
+
b
{\displaystyle \ y-y_{o}=m(x-x_{o})+b}
4. The slope that we have discovered using the differentiated equation can be substituted from m
y
−
y
o
=
17
(
x
−
x
o
)
+
b
{\displaystyle \ y-y_{o}=17(x-x_{o})+b}
5. Now, we must use the given x and y co-ordinates (is y or x is not not given, use the original equation to determine their value) and substitute them into the x and y positions of our straight line equation.
y
−
0
=
17
(
x
−
1
)
+
b
{\displaystyle \ y-0=17(x-1)+b}
y
=
17
x
−
17
+
b
{\displaystyle \ y=17x-17+b}
6. Finally, solve for b and we have the straight line equation
0
=
17
(
1
)
−
17
+
b
{\displaystyle \ 0=17(1)-17+b}
b
=
0
{\displaystyle \ b=0}
Solution:
y
=
17
x
−
17
{\displaystyle \ y=17x-17}
