A-level Mathematics/OCR/C3/Differentiation

Currently, the only functions that you are able to differentiate are relatively simple ones. We will now have a look at some more complicated functions and some new techniques to differentiate them.

The natural functions have two special derivative that you need to memorize. The derivative ${\displaystyle \operatorname {e} ^{x}}$ is unique because its derivative is itself. If there is a constant in the index then the derivative will be multiplied by the constant. The derivative of ${\displaystyle \ln x\,}$ is important because any constant in front of x will be dropped in the derivative.

${\displaystyle {\frac {d}{dx}}\operatorname {e} ^{kx}=k\operatorname {e} ^{kx}}$

${\displaystyle {\frac {d}{dx}}\ln x={\frac {1}{x}}}$

The derivative of an exponential function to any base is: ${\displaystyle {\frac {d}{dx}}\operatorname {a} ^{x}=\operatorname {a} ^{x}\ln(\operatorname {a} )}$

Find the derivative of ${\displaystyle y=\operatorname {e} ^{6x}+\ln 5x}$.

Using the rules we get:

${\displaystyle {\frac {dy}{dx}}=6\operatorname {e} ^{6x}+{\frac {1}{x}}}$

Remember: ${\displaystyle \ln 5x=\ln 5+\ln x\,}$, so ${\displaystyle {\frac {d}{dx}}\ln 5x={\frac {d}{dx}}\ln 5+\ln x=0+{\frac {1}{x}}}$.

The Product Rule states if ${\displaystyle y=f\left(x\right)\cdot g\left(x\right)\,}$ then ${\displaystyle {\frac {dy}{dx}}=f'\left(x\right)g\left(x\right)+f\left(x\right)g'\left(x\right)}$.

Find the derivative of ${\displaystyle y=x^{2}\cdot 18x}$

${\displaystyle {\frac {dy}{dx}}=2x\cdot 18x+x^{2}\cdot 18}$

${\displaystyle {\frac {dy}{dx}}=36x^{2}+18x^{2}}$

${\displaystyle {\frac {dy}{dx}}=54x^{2}}$

The Quotient Rule states: If ${\displaystyle y={\frac {f(x)}{g(x)}}}$ then ${\displaystyle {\frac {dy}{dx}}={\frac {f^{'}(x)g(x)-g^{'}(x)f(x)}{\left\{g(x)\right\}^{2}}}}$.

Find the derivative of ${\displaystyle y={\frac {5x+3}{2x-1}}}$.

${\displaystyle {\frac {dy}{dx}}={\frac {5\times \left(2x-1\right)-\left(5x+3\right)\times 2}{\left\{2x-1\right\}^{2}}}}$

${\displaystyle {\frac {dy}{dx}}={\frac {10x-5-10x-6}{\left\{2x-1\right\}^{2}}}}$

${\displaystyle {\frac {dy}{dx}}={\frac {-11}{\left\{2x-1\right\}^{2}}}}$

The Chain Rule states: If ${\displaystyle y=f[g(x)]\,}$ then ${\displaystyle {\frac {dy}{dx}}=f^{'}[g(x)]\cdot g^{'}(x)}$.

Find the derivative of ${\displaystyle y=\left(2x^{2}-3\right)^{2}}$

${\displaystyle {\frac {dy}{dx}}=2\left(2x^{2}-3\right)4x}$

${\displaystyle {\frac {dy}{dx}}=8x\left(2x^{2}-3\right)}$

If we want the derivative of the inverse function we need to use the following rule: ${\displaystyle {\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}\ if\ {\frac {dx}{dy}}\neq 0}$.

What is the derivative of the inverse of ${\displaystyle f\left(x\right)={\frac {x-1}{x+2}}}$ ?

Find the derivative of the original function using the quotient rule.

${\displaystyle {\frac {dy}{dx}}={\frac {3}{(x+2)^{2}}}}$

To find the derivative of the inverse we use the rule.

${\displaystyle {\frac {dx}{dy}}={\frac {(x+2)^{2}}{3}}}$

Differentiation can be used for situations where you need to find the maximum or minimum value of a problem depending on variables that affect each other. Problems involving connected rates of changes are very diverse and are used in a lot of practical applications such as find the maximum volume of a box from the least amount of cardboard to calculating how long it will take a container to become filled. You will need to write out the problems algebraically and then use the first derivatives to find the maximum or minimum values. The general steps are:

1. The first step is listing what we know and what shape we are dealing with.
2. The second second step is to figure what we need to solve for.
   1. If the quantities are related solve for one variable in terms of the other.
3. The third step is the set up the equation and differentiate with respect to the variable.
4. Then input all the relevant knows into the equation.
5. If any data is missing see if you can obtain it from an equation.
6. Finish the problem.
7. Write the answer.

If a moving ladder (C) 6 metres long is leaning against a wall and the distance between the wall and the bottom of the ladder increases at a rate of 1.5 metres per second how fast is the top of the ladder moving down the wall when the bottom of the ladder is 3 metres from the wall?

1. The first step is listing what we know and what shape we are dealing with.
   1. The shape is a triangle. So we will be using ${\displaystyle a^{2}+b^{2}=c^{2}}$.
   2. A = ? B = 3 ${\displaystyle {\frac {db}{dt}}=1.5}$ C = 6
2. The second step is to figure what we need to solve for.
   1. ${\displaystyle {\frac {da}{dt}}}$
3. The third step is the set up the equation and differentiate with respect to the variable. Which is time in this equation.

1. 1. ${\displaystyle 2a{\frac {da}{dt}}+2b{\frac {db}{dt}}=0{\frac {dc}{dt}}}$
2. Then input all the relevant knows into the equation.
   1. ${\displaystyle a{\frac {da}{dt}}+3\times 1.5=0}$
3. If any data is missing see if you can obtain it from an equation.
   1. ${\displaystyle a^{2}+3^{2}=6^{2}}$
   2. ${\displaystyle a=3{\sqrt {3}}}$
4. Finish the problem.
   1. ${\displaystyle 3{\sqrt {3}}{\frac {da}{dt}}+3\times 1.5=0}$
   2. ${\displaystyle {\frac {da}{dt}}=-{\frac {4.5}{3{\sqrt {3}}}}}$
5. When the bottom of the ladder is 3 metres away from the wall the top will be sliding down at a rate of ${\displaystyle -{\frac {4.5}{3{\sqrt {3}}}}{\frac {metres}{s}}={\frac {\sqrt {3}}{2}}{\frac {metres}{s}}}$.

If sand is being poured on the ground in a conical shape at a rate of 5 cubic centimetres a minute. The sand is piling up such that the radius of the pile is equal to the height of the pile. How fast is the radius growing when the sand pile has a volume of 9π cubic centimetres?

1. The first step is listing what we know and what shape we are dealing with.
   1. The Shape is a cone. The Formula for the volume of a cone is ${\displaystyle v={\frac {1}{3}}\pi r^{2}h}$.
   2. V = 9π, ${\displaystyle {\frac {dv}{dt}}=5}$, r = h
2. The second step is to figure what we need to solve for.
   1. ${\displaystyle {\frac {dr}{dt}}}$
3. The third step is to set up the equation and differentiate with respect to time.
   1. ${\displaystyle v={\frac {1}{3}}\pi r^{2}h}$
   2. Since r = h we have ${\displaystyle v={\frac {1}{3}}\pi r^{3}}$
   3. ${\displaystyle {\frac {dv}{dt}}=\pi r^{2}{\frac {dr}{dt}}}$
4. Then input all the relevant knowns into the equation.
   1. ${\displaystyle 5=\pi r^{2}{\frac {dr}{dt}}}$
5. If any data is missing see if you can obtain it from an equation.
   1. ${\displaystyle 9\pi ={\frac {1}{3}}\pi r^{3}}$
   2. ${\displaystyle r=3}$
6. Finish the problem.
   1. ${\displaystyle 5=\pi 3^{2}{\frac {dr}{dt}}}$
   2. ${\displaystyle {\frac {dr}{dt}}={\frac {5}{9\pi }}}$
7. When the volume is 9π ${\displaystyle cm^{3}}$ the radius will be growing at a rate of ${\displaystyle {\frac {5}{9\pi }}{\frac {cm}{min}}}$.

Given a rope of 5 meters, what is the biggest area that it can engulf if it is shaped into a rectangle?

What have we got?

${\displaystyle x+x+y+y=5}$		(1)		all sides added equals 5
${\displaystyle a=x*y}$		(2)		a is the area of the rectangle

Finding an expression for ${\displaystyle y}$:

${\displaystyle 2x+2y=5}$
${\displaystyle x+y={\frac {5}{2}}}$
${\displaystyle y={\frac {5}{2}}-x}$

Finding an expression for the area that only depends on ${\displaystyle x}$:

${\displaystyle a=x*y}$
${\displaystyle a=x*({\frac {5}{2}}-x)}$
${\displaystyle a=x*{\frac {5}{2}}-x^{2}}$

Now, let ${\displaystyle f(x)}$ be the function for the area of the rectangle:

${\displaystyle f(x)={\frac {5}{2}}*x-x^{2}}$

Find the derivative, ${\displaystyle f'(x)}$:

${\displaystyle f'(x)={\frac {5}{2}}-2x}$

Finding ${\displaystyle y}$ when ${\displaystyle x}$ is ${\displaystyle {\frac {5}{4}}}$

${\displaystyle 2x+2y=5}$

(1)

${\displaystyle 2*{\frac {5}{4}}+2y=5}$

${\displaystyle 2y=5-{\frac {5}{2}}}$

${\displaystyle 2y={\frac {5}{2}}}$

${\displaystyle y={\frac {5}{2}}*{\frac {1}{2}}}$

${\displaystyle y={\frac {5}{4}}}$

So, the optimal size of the rectangle is ${\displaystyle {\frac {25}{16}}m^{2}}$, with the length of all sides at ${\displaystyle {\frac {5}{4}}m}$.