Notes - Matlan - Derivatives

General Formulae

ƒ(x)±g(x)

(ƒ^′)(x)±(g^′)(x)

ƒ(x)⋅g(x)

(ƒ^′)(x)*g(x)+ƒ(x)*(g^′)(x)

ƒ(x)*g(x)*h(x)

(ƒ^′)(x)*g(x)*h(x)+ƒ(x)*(g^′)(x)*h(x)+ƒ(x)*g(x)*(h^′)(x)

ƒ(x)/g(x)=((ƒ^′)(x)*g(x)-ƒ(x)*(g^′)(x))/(g(x)2)

ƒ(x)n

n⋅ƒ(x)(n-1)⋅(ƒ^′)(x)

ƒ(x)∘g(x)

(ƒ^′)(g(x))⋅(g^′)(x)

Trigonometric Functions

sin(ƒ(x))

cos(ƒ(x))⋅(ƒ^′)(x)

cos(ƒ(x))

-sin(ƒ(x))⋅(ƒ^′)(x)

tan(ƒ(x))

sec2(ƒ(x))⋅(ƒ^′)(x)

sec(ƒ(x))

sec(ƒ(x))⋅tan(ƒ(x))⋅(ƒ^′)(x)

cot(ƒ(x))

-csc2(ƒ(x))⋅(ƒ^′)(x)

csc(ƒ(x))

-csc(ƒ(x))⋅cot(ƒ(x))⋅(ƒ^′)(x)

Some Examples

Example 1

y=4*x2+3*sin(x)-cos(x)

(y^′)=8*x+3*cos(x)+sin(x)

Example 2

y=sin(6*x)

(y^′)=cos(6*x)⋅6

Example 3

y=cos(2*x2-6*x+1)

(y^′)=-sin(2*x2-6*x+1)⋅(4*x-6)

Example 4

y=5*sin(3*x+1)-1/2*cos(4*x+3)

(y^′)=5⋅3⋅cos(3*x+1)-1/2⋅4⋅(-sin(4*x+3))

(y^′)=15*cos(3*x+1)+2*sin(4*x+3)

Example 5

y=x2*sin(3*x)

u=x2

(u^′)=2*x

v=sin(3*x)

(v^′)=3*cos(3*x)

(y^′)=x2⋅3*cos(3*x)+2*x⋅sin(3*x)=3*x2*cos(3*x)+2*x*sin(3*x)

Example 6

y=(1+cos(x))/(-sin(x))

u=1+cos(x)

(u^′)=-sin(x)

v=-sin(x)

(v^′)=-cos(x)

(y^′)=(-sin(x)⋅(-sin(x))-(1+cos(x))⋅(-cos(x)))/((-sin(x))2)

(y^′)=((sin^2)(x)+cos(x)+(cos^2)(x))/(sin^2)(x)

(y^′)=(1+cos(x))/(sin^2)(x)

(sin^2)(x)=1-(cos^2)(x)=(1+cos(x))*(1-cos(x))

(y^′)=(1+cos(x))/((1+cos(x))*(1-cos(x)))=1/(1-cos(x))

Example 7

y=2*(tan^3)(1-2*x)

(y^′)=2⋅3⋅(tan^2)(1-2*x)⋅(sec^2)(1-2*x)⋅(-2)

(y^′)=-12⋅(tan^2)(1-2*x)⋅(sec^2)(1-2*x)

Example 8

y=16*cot4(2*x2-5*x+2)

(y^′)=16⋅4⋅(cot^3)(2*x2-5*x+2)⋅(-(csc^2)(2*x2-5*x+2)⋅(4*x-5))

(y^′)=-64⋅(4*x-5)⋅(cot^3)(2*x2-5*x+2)⋅(csc^2)(2*x2-5*x+2)

Example 9

y=3*(sec^4)(2*x3-5*x)

(y^′)=3⋅4⋅sec3(2*x3-5*x)⋅tan(2*x3-5*x)⋅sec(2*x3-5*x)⋅(6*x2-5)

(y^′)=12⋅(6*x2-5)⋅(sec^4)(2*x3-5*x)⋅tan(2*x3-5*x)

Applying Derivatives

Gradient of Tangents

If you want to find the gradient of a curve's tangent at a certain point you can just find the derivative at that point, because the derivative is the rate of change.

Recall the following formula to find a straight line's gradient.

m=Δ(y)/Δ(x)=(ƒ((x_2))-ƒ((x_1)))/((x_2)-(x_1))

As the difference between (x_2) and (x_1) gets smaller, the gradient eventually becomes the derivative at that point.

m=(lim_h→0)((ƒ((x_1)+h)-ƒ((x_1)))/h)

That's literally just a derivative.

m=(ƒ^′)((x_1))=d(y)/d(x)

Example 1

Find the tangent of y=x2+3*x+4 at x=2

(y^′)=2*x+3

m=2⋅2+3=7

Therefore, the tangent is:

y=7*x+c

We need to find c by first finding the y coordinate of the tangent point.

y=22+3⋅2+4=14

Then, we need to find c. I use this method because it's easier to think about. Simply substitute to the incomplete form of the tangent line.

14=7⋅2+c

c=0

Therefore, the equation of the tangent is y=7*x

Next, what if we want to find the normal?

The normal's gradient has the negative inverse of the gradient of the tangent.

m=-1/(m^_tangent)

m=-1/7

Therefore, the equation of the normal is the following.

y=-1/7*x+c

Then, we substitute x=2 and y=14 once again.

14=-2/7+c

c=100/7

Finally, we find the equation of the normal.

y=-x/7+100/7

x+7*y-100=0

Example 2

Find the tangent of y=(x2)/4+x-4 that is parallel to 6*x-2*y-5=0

Okay, first let's find the gradient we need.

2*y=6*x-5

y=-5/2+3*x

Therefore, the gradient is m=3

Next, we need to find the derivative of the curve.

(y^′)=1/4⋅2⋅x+1=x/2+1

Then, we need to substitute (y^′) for m=3

3=x/2+1

x=4

We found the x coordinate where the tangent point will be. Now, let's find the y coordinate by substituting into the original curve.

y=(42)/4+4-4=4

Therefore, the tangent point is at (4,4). We now need to find the equation for the tangent.

y=m*x+c

y=3*x+c

Substitute x=4 and y=4

4=3⋅4+c

c=-8

y=3*x-8

Found it!

Example 3

Find the tangent of the curve y=x2-4*x+3 that is perpendicular to the line 4*x+8*y-1=0

Well, let's find the gradient of the given line first.

8*y=-4*x+1

y=-x/2+1/8

m=-1/2

Therefore, the gradient of the tangent is m=2

Then, we need to find where the tangent is located by first deriving the original curve.

(y^′)=2*x-4

2=2*x-4

x=3

We found x=3 to be the x coordinate of the tangent point. Now, let's find the y coordinate.

y=32-4⋅3+3=0

So, the tangent point is at (3,0)

Finally, let's solve for the equation of the tangent.

y=m*x+c

y=2*x+c

0=3⋅2+c

c=-6

y=2*x-6

Simply lovely.

Example 4

Find the coordinate of the tangent point if y=2*x is a tangent of y=x2+4*x+1

First, let's derive the equation of the curve.

(y^′)=2*x+4

We know that the gradient of the tangent is m=2

2=2*x+4

x=-1

We find that the x coordinate of the tangent point is x=-1, now let's find the y coordinate.

y=(-1)2+4⋅(-1)+1=-2

Therefore, the point of tangency is at (-1,-2)

Well, technically you can just substitute y=2*x into the original equation and find the solution to that, since all parabolic tangents have only 1 intersection point each.

2*x=x2+4*x+1

x2+2*x+1=0

(x+1)2=0

x=-1

And so on. It's probably easier this way.

Example 5

The curve y=(x2-a)/x has a tangent of y=b*x-2 at x=1

Find a.

y=x-a/x

Substitute x=1 into both equations.

(1-a)/1=b-2

1-a=b-2

a=3-b

b=3-a

(y^′)=1+a*x(-2)

Substitute (y^′)=b and x=1

b=1+a

b=3-a

2*b=4

b=2

2=1+a

a=1

Example 6

The line y=a*x+b is perpendicular with y=2*x+7, ∴a=2

The line is also a tangent of y=x√(,x)-x

Find a+b

Let's first find the derivative.

y=x(3/2)-x

(y^′)=3/2*x(1/2)-1

2=3/2√(,x)-1

3/2√(,x)=3

9/4*x=9

∴x=4

We found the x coordinate of the tangent point.

y=2*x+b

y=x(3/2)-x

Those 2 are supposed to interesect at x=4

y=8+b

y=4(3/2)-4=4

4=8+b

b=-4

a=2

a+b=2-4=-2

Easy.

Increasing and Decreasing Functions

Increasing functions have a positive gradient, while decreasing functions have a negative gradient. Stationary functions have a gradient of 0.

The gradient of a function at x can be found via the derivative of the function.

Between 2 values of a first derivative that are positive and negative, there has to be an inflection point. The inflection point of a function can be found wherever (ƒ^′*′)(x)=0. The inflection point isn't the same as extrema, rather it's where the curve starts curving in the other direction.

The first derivative is for finding the gradient of the function.

The second derivative is for finding the properties of the function's curvature.

• Increasing: (ƒ^′)(x)>0

• Decreasing: (ƒ^′)(x)<0

• Stationary: (ƒ^′)(x)=0

• Inflection point: (ƒ^′*′)(x)=0

• Concave (opens) upwards: (ƒ^′*′)(x)>0

• Concave (opens) downwards: (ƒ^′*′)(x)<0

Example 1

Find the interval of increasing point, decreasing point, stationary point, inflection point, and where it curves upwards and downwards.

ƒ(x)=1/3*x3-2*x2-21*x

(ƒ^′)(x)=x2-4*x-21

Therefore, the stationary point is wherever (ƒ^′)(x)=0

x2-4*x-21=0

x={-3,+7}

And just substitute it.

ƒ(-3)=1/3⋅(-3)3-2⋅(-3)2-21⋅(-3)=36

ƒ(7)=1/3⋅73-2⋅72-21⋅7=-392/3

Therefore, the stationary points are at (-3,36) and (7,-392/3)

Okay. We know the stationary points now. We can use that to make a table

To determine whether it's + - + or the inverse, simply use a test point, say x=0 with the first derivative.

Now we know where the function increases and decreases.

Next, let's find the inflection point.

(ƒ^′*′)(x)=0=2*x-4

x=2

I'm not actually gonna find the inflection point because it's just ƒ(2)

Now that we know the inflection point is at x=2, using the table from before, we can conclude 2 things:

• The function is concave downwards at x<2

• The function is concave upwards at x>2

Example 2

A box without a cover (5 sides) has a base that is in the shape of a square is going to be made with a piece of paper. If the area of the box that can be made with the paper is 4800, find the size of the box so the volume is maximized.

V=p⋅l⋅t=p2*t

That's the target.

4800=p2+4*p*t

p2+4*p*t-4800=0

That's the constraint.

How do you find the target via the constraint?

Just subtitute t

4*p*t=4800-p2

t=(4800-p2)/(4*p)

Substitute that to V=p2*t

V=p2⋅(4800-p2)/(4*p)=1200*p-(p3)/4

We need to find the maximum via deriving it.

d(V)/d(p)=-3/4*p2+1200

We find the maximum via finding the stationary points, then testing each one.

-(3*p2)/4+1200=0

p={-40,+40}

In this case, there is only 1 stationary point, as it's impossible to have a negative length.

∴p=40

∴t=(4800-1600)/160=20

∴V=1600⋅20=32000

It's easy.

Example 3

Ali has 80of wire that he will use to make a cage that is in the shape of a rectangle, which has one side facing a warehouse. The side facing the warehouse isn't fenced. Find the area of land that can be fenced.

A=p⋅l

Assume one the shortest sides will be the one facing the warehouse.

2*p+l=80

l=80-2*p

A=p⋅(80-2*p)=-2*p2+80*p

d(A)/d(p)=-4*p+80

-4*p+80=0

p=20

2⋅20+l=80

l=40

A=20⋅40=800

Example

A rectangle is located inside half a circle (assume the graph is symmetric and centered at x=0) that has a radius of 4. Find the maximum area of the rectangle.

A=p⋅l

Obviously, the diagonal of the circle has to be equal to 4

42=(p/2)2+l2

16=(p2)/4+l2

(p2)/4=16-l2

p2=64-4*l2

p=√(,64-4*l2)

A=√(,64-4*l2)⋅l

d(A)/d(l)=(-8*l2+64)/√(,64-4*l2)

-8*l2+64=0

l={-2√(,2),+2√(,2)}

p=√(,64-4⋅(2√(,2))2)=4√(,2)

A=2√(,2)⋅4√(,2)=16

Example 5

The price to produce a radio of x amounts is 1/4*x2+35*x+25

Each radio is sold at 50-1/2*x

Find the amount of radius sold where you get the most money.

So, the total amount you get from selling x radios is the following.

50*x-1/2*x2

Therefore, we can find the profit P by subtracting.

1/4*x2+35*x+25-50*x+(x2)/2=25-15*x+3/4*x2=P

Then, we find the stationary points.

d(P)/d(x)=3/2*x-15=0

3/2*x-15=0

3/2*x=15

x=10

Example 5

Find the intervals of increasing and decreasing

Find the stationary points

Find the inflection points

y=4*cos(x)+cos(2*x)

0<x<360

y=4*cos(x)+2*(cos^2)(x)-1

(y^′)=-4*sin(x)+4*cos(x)*(-sin(x))

(y^′)=-4*sin(x)*(1+cos(x))

Let's find the increasing and decreasing intervals.

-4*sin(x)*(1+cos(x))=0

1+cos(x)=0

x=π

-4*sin(x)=0

x={0,+π}

∴x=π

y=4*cos(π)+cos(2*π)=-3

Therefore, the stationary point is at (π,-3)

Let's check a random point.

(ƒ^′)(π/2)=-4*sin(π/2)*(1+cos(π/2))=-4

So, the function increases at x>180and decreases at x<180

Let's now find the inflection point.

(d^2)(y)/d(x2)=-4*cos(x)-4*cos(2*x)

-4*cos(x)-4*cos(2*x)=0

cos(x)+cos(2*x)=0

cos(x)+2*(cos^2)(x)-1=0

2*(cos^2)(x)+cos(x)-1=0

Well that's a quadratic.

(2*cos(x)-1)*(2*cos(x)+2)=0

2*cos(x)-1=0

x={π/3,+5/3*π}

2*cos(x)+2=0

x=π

So, the inflection points are at π/3,π,and 5/3*π