Featured 2 weeks ago

The **Taylor polynomial** is just another name for the full Taylor series truncated at a finite

Some common errors are:

- Letting
#x = a# within the#(x - a)^n# term. - Taking the derivatives at the same time as writing out the series, which is not necessarily wrong, but in my opinion, it allows more room for silly mistakes.
- Not plugging
#a# in for the#n# th derivative, or plugging in#0# .

The formula to write out the series was:

#sum_(n=0)^(oo) (f^((n))(a))/(n!)(x-a)^n#

So, we would have to take some derivatives of

#f^((0))(x) = f(x) = x/(1 + x)#

#f'(x) = x*(-(1 + x)^(-2)) + 1/(1+x)#

#= -x/(1 + x)^2 + 1/(1+x)#

#= -x/(1 + x)^2 + (1 + x)/(1+x)^2#

#= 1/(1+x)^2#

#f''(x) = -2(1 + x)^3 = -2/(1+x)^3#

#f'''(x) = 6/(1 + x)^4#

#f''''(x) = -24/(1 + x)^5#

etc.

So plugging things in gives (truncated at

#=> (f^((0))(a))/(0!)(x-a)^0 + (f'(a))/(1!)(x-a)^1 + (f''(a))/(2!)(x-a)^2 + (f'''(a))/(3!)(x-a)^3 + (f''''(a))/(4!)(x-a)^4 + . . . #

#= a/(1 + a) + 1/(1+a)^2(x-a) + (-2/(1+a)^3)/(2)(x-a)^2 + (6/(1+a)^4)/(6)(x-a)^3 + (-24/(1 + a)^5)/(24)(x-a)^4 + . . . #

#= color(blue)(a/(1 + a) + 1/(1+a)^2(x-a) - 1/(1+a)^3(x-a)^2 + 1/(1+a)^4(x-a)^3 - 1/(1 + a)^5(x-a)^4 + . . . )#

Featured 4 weeks ago

The derivative is given (implicitly) by;

# x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

The equation of the tangent at

# y = x + 1/4 #

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

When we differentiate

However, we cannot differentiate a non implicit function of

When this is done in situ it is known as implicit differentiation.

We have:

# x^2 + y^2 = (5x^2 + 4y^2 -x)^2 #

Differentiate wrt

# 2x+2ydy/dx = 2(5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

# :. x + ydy/dx = (5x^2 + 4y^2 -x)(10x+8ydy/dx-1) #

Whilst we could spend time and effort to find an explicit equation for

# => 0 + 1/4 dy/dx = (0 + 4(1/16) -0)(0+8/4dy/dx-1) #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = (1/4)(2dy/dx-1) #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/2dy/dx-1/4 #

# :. \ \ \ \ \ \ \ \ 1/4 dy/dx = 1/4 #

# :. \ \ \ \ \ \ \ \ \ \ \ \ dy/dx = 1 #

So the tangent passes through

# \ \ \ \ \ y-1/4 = 1(x-0) #

# :. y-1/4 = x #

# :. \ \ \ \ \ \ \ \ y = x + 1/4 #

We can verify this solution graphically;

**Advanced Calculus**

There is another (often faster) approach using partial derivatives. Suppose we cannot find

# (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = âˆ’((partial F)/(partial x)) / ((partial F)/(partial y)) #

So Let

#(partial F)/(partial x) = 2x - 2(5x^2 + 4y^2 -x)(10x-1)#

#(partial F)/(partial y) = 2y - 2(5x^2 + 4y^2 -x)(8y) #

And so:

# dy/dx = -(2x - 2(5x^2 + 4y^2 -x)(10x-1))/(2y - 2(5x^2 + 4y^2 -x)(8y))#

# \ \ \ \ \ \= -(x - (5x^2 + 4y^2 -x)(10x-1))/(y - 8y(5x^2 + 4y^2 -x))#

Here we get an immediate implicit function for the derivative, so again at

# dy/dx = -(0 - (0 + 4/16 -0)(0-1))/(1/4 - 8/4(0 + 4/16 -0))#

# \ \ \ \ \ \ = -(- (1/4)(-1))/(1/4 - 2(1/4))#

# \ \ \ \ \ \ = -(1/4)/(-1/4)#

# \ \ \ \ \ \ = 1 \ \ \ \ # , as before

Featured 3 weeks ago

The idea here is that you can find the **rate of change** of the pollution with respect to time by taking the first derivative of your function

So this is pretty much an exercise in finding the derivative of the function

#P(t) = (t^(1/4) + 3)^3#

which can be calculated by using the chain rule and the power rule. Keep in mind that you have

#color(blue)(ul(color(black)(d/(dt)[u(t)^n] = n * u(t)^(n-1) * d/(dt)[u(t)])))#

In your case,

#{(u(t) = t^(1/4) + 3), (n = 3) :}#

This means that the derivate of

#overbrace(d/d(dt)[P(t)])^(color(blue)(=P^(')(t))) = 3 * (t^(1/4) + 3)^2 * d/(dt)(t^(1/4) + 3)#

#P^'(t) = 3 * (t^(1/4) + 3)^2 * 1/4 * t^((1/4 - 1))#

#P^'(t) = 3/4 * t^(-3/4) * (t^(1/4) + 3)^2#

Now all you have to do is plug in

#P^'(t) = 3/4 * 16^(-3/4) * (16^(1/4) + 3)^2#

Since you know that

#16 = 2^4#

you can rewrite the equation as

#P^'(16) = 3/4 * 2^[[4 * (-3/4)]] * [2^((4 * 1/4)) + 3]^2#

#P^'(16) = 3/4 * 2^(-3) * (2 + 3)^2#

#P^'(16) = 3/4 * 1/8 * 25#

#P^'(16) = 75/32#

And there you have it -- the rate at which the pollution changes after

Featured 3 weeks ago

Recall the formula for chain rule:

#color(blue)(bar(ul(|color(white)(a/a)dy/dx=dy/(du)(du)/dxcolor(white)(a/a)|)))# or#color(blue)(bar(ul(|color(white)(a/a)f'(x)=g'[h(x)]h'(x)color(white)(a/a)|)))#

and the formula for power rule:

#color(blue)(bar(ul(|color(white)(a/a)d/dx(x^n)=nx^(n-1)color(white)(a/a)|)))#

To start, recognize the inside and outside functions of

Inside function:

#y=color(darkorange)(4+3x)#

Outside function:#y=color(green)(sqrt(a))#

**How to Differentiate Using Chain Rule**

#1# . Take the derivative of the outside function,#y=sqrt(a)# , but replace the#a# with the inside function,#4+3x# .

#2# . Multiply by the derivative of the inside function,#4+3x# .

**Applying Chain Rule**

1. The derivative of the outside function,

#y=sqrt(a)#

#color(red)(darr)#

#y=1/2a^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)#

2.

#y=1/2(4+3x)^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)(3)#

#color(green)(bar(ul(|color(white)(a/a)y=3/(2(4+3x)^(1/2))color(white)(a/a)|)))#

Featured 3 weeks ago

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1 #

**Method 1 : Graphically**

graph{(e^(2x)-e^x)/(x) [-8.594, 9.18, -1.39, 7.494]}

Although far from conclusive, it appears that:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = 1#

**Method 2 : L'HÃ´pital's rule**

The limit:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) #

is of an indeterminate form

# lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x)) #

And so applying L'HÃ´pital's rule we get:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (2e^(2x)-e^x)/(1)#

# " "= 2-1#

# " "= 1#

**Method 3 - Power Series**

The power series for

# e^x = 1 + x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #

And so we have:

# e^(2x) = 1 + 2x + (2x)^2/(2!) + (2x)^3/(3!) + (2x)^4/(4!) + ... #

Therefore;

# (e^(2x)-e^x)/(x) = { (1 + 2x + O(x^2)) - (1 + x + O(x^2)) } / x #

# " " = { 1 + 2x + O(x^2) - 1 - x + O(x^2) } / x #

# " " = { x + O(x^2) } / x #

# " " = 1 + O(x) #

And so:

# lim_(x rarr 0) (e^(2x)-e^x)/(x) = lim_(x rarr 0) (1 + O(x)) #

# " " = 1 #

Featured 2 weeks ago

C)

Let's first imagine what the region would look like. Inverse functions are reflections of themselves over the line

We see that the top line will be given by

The inverse of this is given by:

#y=9/25x+b" "=>" "x=9/25y+b#

#color(white)(y=9/25x+b)" "=>" "y=25/9(x-b)=f^-1(x)#

Since this is the lower line we say that

The area of the region is given by two distinct parts: the first is the trapezoid that lies under

The second part is a triangle that extends from the

Let's find these important points first:

**-intercept**

#L_2(x)=9/25(x-b)=0" "=>" "x=b#

**Intersection Point**

#L_1(x)=L_2(x)" "=>" "9/25x+b=25/9(x-b)#

#color(white)(L_1(x)=L_2(x))" "=>" "81/625x+9/25b=x-b#

#color(white)(L_1(x)=L_2(x))" "=>" "34/25b=544/625x#

#color(white)(L_1(x)=L_2(x))" "=>" "x=25/16b#

We can now, without any integration necessary (since we're dealing with lines) find these areas.

**Trapezoid Area**

One base is

#A_1=1/2b(b+34/25b)=59/50b^2#

**Triangle Area**

The base is again

#A_2=1/2(34/25b)(9/16b)=153/400b^2#

The total region

#R=A_1+A_2=59/50b^2+153/400b^2=49#

#25/16b^2=49#

#b=sqrt((49(16))/25)=28/5#

Featured 2 weeks ago

At points

Slope of a tangent at a point on the curve

i.e.

=

and as we have to identify, where slope of tangent is

and as

or

or

i.e.

we have slope of tangent as

graph{(x^2y^2+xy-2)(x-y)=0 [-10, 10, -5, 5]}

Featured 2 weeks ago

Absolute maximum:

Absolute minimum:

We start by differentiating.

#p'(x) = 3x^2 - 18x#

We now find the critical numbers, which occur when the derivative is

#0 = 3x(x - 6)#

#x= 0 and 6#

We now test around the points to check whether the derivative is increasing or decreasing.

**Test Point: #x = -1#**

#p(-1) = 3(-1)^2 - 18(-1) = 3 + 18 = 21#

This means that

#p(-2) = (-2)^3 - 9(-2)^2 + 10#

#p(-2) = -8 -9(4) +10#

#p(-2) = -34#

This works!

#p(8) = 8^3 - 9(8)^2 + 10#

#p(8) = 512 - 576 + 10#

#p(8) = -54#

This works!

Therefore,

Finally, we must ensure that

**Test Point #x = 5#**

#p(5) = 5^3 - 9(5)^2 + 10#

#p(5) = -90#

We can now see that

#p(6)= 6^3 - 9(6)^2 + 10#

#p(6) = -98#

Since this is smaller than both

Hopefully this helps!

Featured 5 days ago

# 2+pi/4 #

Here is the graph of the two curves. The shaded area,

It is a symmetrical problems so we only need find the shaded area of the RHS of Quadrant

We could find the angle

# r=1+cos 2theta #

# r=1 #

However, intuition is faster, and it looks like angle of intersection in

# theta = pi/4 => r=1+cos 2theta =1+cos(pi/2) = 1 #

Confirming our intuition. So now we have the following:

Where the shaded area repents

We can now start to set up a double integral to calculate this area

#r# sweeps out a ray from#1# to#1+cos(2theta)#

#theta# varies for#0# to#pi/4#

So then:

# 1/4A = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #

If we evaluate the inner integral first then we get:

# int_1^(1+cos2theta) r \ dr = [1/2r^2]_1^(1+cos2theta) #

# " " = 1/2( (1+cos2theta)^2 - 1^2) #

# " " = 1/2( 1+2cos2theta+cos^2 2theta - 1) #

# " " = 1/2( 2cos2theta+cos^2 2theta) #

# " " = 1/2( 2cos2theta+1/2(cos4theta+1)) #

# " " = cos2theta+1/4cos4theta+1/4 #

And so our double integral becomes:

# 1/4A \ = int_0^(pi/4) int_1^(1+cos2theta) r \ dr \ d theta #

# " " = int_0^(pi/4) cos2theta+1/4cos4theta+1/4 \ d theta #

# " " = [1/2sin2theta+1/16sin4theta+1/4theta]_0^(pi/4) #

# " " = (1/2sin(pi/2)+1/16sinpi+1/4pi/4) - (0) #

# " " = 1/2+pi/16 #

# :. A = 2+pi/4 #

# #

**METHOD 2**

If you are not happy with double integrals, then we can evaluate the area using the polar area formula

Here this shaded area is given by:

# A_1 = 1/2 \ int_0^(pi/4) \ (1+cos 2theta)^2 \ d theta #

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+ cos^2 2theta \ d theta #

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2(cos4theta+1) \ d theta#

# \ \ \ \ = 1/2 \ int_0^(pi/4) \ 1+2cos 2theta+1/2cos4theta+1/2 \ d theta#

# \ \ \ \ = 1/2 [3/2theta+sin2theta+1/8sin4theta]_0^(pi/4)#

# \ \ \ \ = 1/2 ((3pi)/8+1+0)#

# \ \ \ \ = (3pi)/16+1/2#

And this shaded area

is given by:

# A_2 = 1/2 \ int_0^(pi/4) \ (1)^2 \ d theta #

# \ \ \ \ = 1/2 [theta]_0^(pi/4)#

# \ \ \ \ = 1/2 (pi/4-0)#

# \ \ \ \ = pi/8#

And so the total sought area is

# 1/4 A \ = A_1 - A_2 #

# " " = (3pi)/16+1/2 - pi/8 #

# " " = (pi)/16+1/2 #

#:. A = (pi)/4+2 #

Featured 4 days ago

Yes, but not just one formula. There are many.

The derivative (of a differentiable function),

# f'(a) = lim_(x rarr a )(f(x)-f(a))/(x-a) #

With a slight change of notation we can write:

# dy/dx = f'(x) = lim_(h rarr 0 ) (f(x+h)-f(x))/h #

In some older texts the notation may involve

# f'(x) = lim_(deltax rarr 0 ) (f(x+deltax)-f(x))/(deltax) = lim_(Deltaxrarr 0 ) (f(x+Deltax)-f(x))/(Deltax)#

It represents both the rate of change of the function, and the gradient of the tangent line at any particular point. If the limit does not exists then the function is not differentiable.

In practice we do not derive the derivative from first principles using the limit definition, but instead we use various rules that can be proved to be true;

Power Rule:#d/dx(x^n) = nx^(n - 1)#

Example: Find

We have

Chain Rule:#d/dx(f(g(x)) = f'(g(x)) * g'(x)# , in other words the derivative of the composition#f(g(x))# is the inner function times the outer function.

Example: Find

We let

Product Rule:#d/dx(f(x) xx g(x)) = f'(x)g(x) + f(x)g'(x)#

Example: Find

We let

Quotient rule:#d/dx((f(x))/(g(x)))=(g(x)f'(x)-f(x)g'(x))/(g(x))^2#

Example: Find

We let

Here are a few other useful derivative formulas I think you should know.

#d/dx(lnx) = 1/x#

#d/dx(sinx) = cosx#

#d/dx(cosx) = -sinx#

Once you get good at the basic differentiation rules, you may be asked to solve problems that combine the differentiation rules in interesting ways.

Example: Find

First of all, note that

There are a few ways of differentiating this.

a) you could use the quotient rule to differentiate

#1/cosx# and then the chain rule to differentiate#y# .b) you could rewrite

#1/cosx# as#(cosx)^-1# and then differentiate using the chain rule twice.c) you could use the laws of logarithms to simplify and then differentiate. I'll use this method.

By the rule

By the chain rule, we have

#dy/dx = -1/cosx * -sinx = sinx/cosx = tanx#

Hopefully you now get a good idea what differentiation is about!