# More Differentiation

## More Differentiation Revision

**More Differentiation**

This section contains** further uses** of the **chain**,** product** and** quotient**** rules**, combining them with **other things** you have learned so far such as **trigonometry** and **stationary points**.

Make sure you are happy with the following topics before continuing.

**Derivatives of sec, cosec and cot**

Use the **quotient rule** to find the **derivatives** of \sec(x), \cosec(x) and \cot(x).

**Recall**: Quotient rule

\dfrac{dy}{dx}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}

**Using the Chain, Product and Quotient Rule With Special Functions**

Now that we know how to **differentiate** every **exponential function** and every **trigonometric function**, we can use them **along with our rules**. Remember when to use each rule:

**Function of a function – chain rule****Multiplying two functions together – product rule****Fraction made of functions – quotient rule**

**Example: **Find the **derivative** of \cosec(4x^{3})

This is a **function of a function**, so use the **chain rule**.

Take t=4x^{3}.

Then y=\cosec(t)

\dfrac{dy}{dt}=-\cosec(t)\cot(t)

\dfrac{dt}{dx}=12x^{2}

Hence: \dfrac{dy}{dx}=-\cosec(t)\cot(t)12x^{2}

Put x back in:

\dfrac{dy}{dx}=-12x^{2}\cosec(4x^{3})\cot(4x^{3})

**Using the Rules Twice**

Sometimes you might need to use the **chain**, **product** or **quotient** **rule** twice to get a **derivative**.

**Example: Differentiate** 2x^{2}\cos(e^{x})

This is a **product of two functions** so use **product rule**.

\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}

u=2x^{2}\;\;\;v=\cos(e^{x})

\dfrac{du}{dx}=4x

To find \dfrac{dv}{dx} we need to use the **chain rule**.

Take t=e^{x}

Then v=\cos(t)

\dfrac{dv}{dt}=-\sin(t)

\dfrac{dt}{dx}=e^{x}

Hence: \dfrac{dv}{dx}=-\sin(t)e^{x}

Put x back in:

\dfrac{dv}{dx}=-e^{x}\sin(e^{x})Now we can complete our **product rule**.

**Using the Rules to find the Second Derivative**

We already know that **second derivatives** can be very useful when it comes to **classifying stationary points**. However, starting with expressions like these, finding the **second derivative** can get **very complicated**.

**Example: **y=e^{3x^{2}}. Find \dfrac{d^{2}y}{dx^{2}}.

To **differentiate once**, we need the **chain rule**, because this is a **function within a function**.

Take t=3x^{2}

Then y=e^{t}

\dfrac{dy}{dt}=e^{t}

\dfrac{dt}{dx}=6x

Hence: \dfrac{dy}{dx}=6xe^{t}

Put x back in:

\dfrac{dy}{dx}=6xe^{3x^{2}}To **differentiate a second time**, we notice we have **two functions multiplied together**, so must use the **product rule**.

\dfrac{d^{2}y}{dx^{2}}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}

u=6x\;\;\;v=e^{3x^{2}}

\dfrac{du}{dx}=6

\dfrac{dv}{dx} requires the **chain rule**, but it is identical to the **differentiation** we just did above.

Hence:

\begin{aligned}\dfrac{d^{2}y}{dx^{2}}&=6x\times6xe^{3x^{2}}+6e^{3x^{2}}\\[1.2em]&=36x^{2}e^{3x^{2}}+6e^{3x^{2}}\\[1.2em]&=6(6x^{2}+1)e^{3x^{2}}\end{aligned}

## More Differentiation Example Questions

**Question 1:**

a) State the derivatives of:

i) \sec(x)

ii) \cosec(x)

iii) \cot(x)

b) Find the derivative of \sec(\cot(x)), stating any rules you use.

**[5 marks]**

a) i) \sec(x)\tan(x)

ii) -\cosec(x)\cot(x)

iii) -\cosec^{2}(x)

b) y=\sec(\cot(x))

This is a function of a function, so use chain rule.

\dfrac{dy}{dx}=\dfrac{dy}{dt}\dfrac{dt}{dx}

Take t=\cot(x)

Then y=\sec(t)

\dfrac{dy}{dt}=\sec(t)\tan(t)

\dfrac{dt}{dx}=-\cosec^{2}(x)

\dfrac{dy}{dx}=-\sec(t)\tan(t)\cosec^{2}(x)

Put x back in

\dfrac{dy}{dx}=-\sec(\cot(x))\tan(\cot(x))\cosec^{2}(x)

**Question 2: **Differentiate \sin(3x^{2})\ln(x)

**[3 marks]**

This is a product of two functions, so use product rule.

\dfrac{dy}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}

u=\sin(3x^{2})\;\;\;v=\ln(x)

\dfrac{dv}{dx}=\dfrac{1}{x}

For \dfrac{du}{dx} we need the chain rule because we have a function within a function.

\dfrac{du}{dx}=\dfrac{du}{dt}\dfrac{dt}{dx}

Take t=3x^{2}

Then y=\sin(t)

\dfrac{du}{dt}=\cos(t)

\dfrac{dt}{dx}=6x

\dfrac{du}{dx}=6x\cos(t)

Put x back in

\dfrac{du}{dx}=6x\cos(3x^{2})

Put into product rule:

\dfrac{dy}{dx}=\dfrac{\sin(3x^{2})}{x}+6x\ln(x)\cos(3x^{2})

**Question 3: **Find and classify the stationary points of y=9e^{\frac{1}{2}x+\cos(x)} where 0<x<\pi

**[8 marks]**

First, find the derivative using the chain rule.

\dfrac{dy}{dx}=9\left(\dfrac{1}{2}-\sin(x)\right)e^{\frac{1}{2}x+cos(x)}

Stationary points are where this is equal to 0.

9\left(\dfrac{1}{2}-\sin(x)\right)e^{\frac{1}{2}x+cos(x)}=0

Exponentials can never be equal to 0, so:

9\left(\dfrac{1}{2}-\sin(x)\right)=0

\dfrac{1}{2}-\sin(x)=0

\sin(x)=\dfrac{1}{2}

x=\dfrac{\pi}{6},\dfrac{5\pi}{6}

Find y-coordinate of stationary points:

y=9e^{\frac{1}{2}\times\frac{\pi}{6}+\cos(\frac{\pi}{6})}

y=9e^{\frac{\pi}{12}+\frac{\sqrt{3}}{2}}

y=9e^{\frac{1}{2}\times\frac{5\pi}{6}+\cos(\frac{5\pi}{6})}

y=9e^{\frac{5\pi}{12}-\frac{\sqrt{3}}{2}}

Stationary points are:

\left(\dfrac{\pi}{6},9e^{\frac{\pi}{12}+\frac{\sqrt{3}}{2}}\right),\left(\dfrac{5\pi}{6},9e^{\frac{5\pi}{12}-\frac{\sqrt{3}}{2}}\right)

Next, differentiate again using the product rule.

u=9\left(\dfrac{1}{2}-\sin(x)\right)

v=e^{\frac{1}{2}x+\cos(x)}

\dfrac{du}{dx}=-9\cos(x)

\dfrac{dv}{dx}=\left(\dfrac{1}{2}-\sin(x)\right)e^{\frac{1}2{x}+\cos(x)}

\dfrac{d^{2}y}{dx^{2}}=9\left(\left(\dfrac{1}{2}-\sin(x)\right)^{2}-\cos(x)\right)e^{\frac{1}{2}x-\cos(x)}

Before substituting in x=\dfrac{\pi}{6} and x=\dfrac{5\pi}{6}, we should first notice that the exponential part of the expression is positive for any x, so we need not keep it when trying to determine the sign of \dfrac{d^{2}y}{dx^{2}}.

Want to determine the sign of 9\left(\left(\dfrac{1}{2}-\sin(x)\right)^{2}-\cos(x)\right)

Secondly, we notice that the x values came from \dfrac{1}{2}-\sin(x)=0, so this part will remain 0 when substituting the x values in. We can also get rid of the factor of 9 without changing the sign.

This means that all we need to do is find the sign of -\cos(x).

-\cos\left(\dfrac{\pi}{6}\right)=-\dfrac{\sqrt{3}}{2}<0 so this is a maximum.

-\cos\left(\dfrac{5\pi}{6}\right)=\dfrac{\sqrt{3}}{2}>0 so this is a minimum.

Hence:

\left(\dfrac{\pi}{6},9e^{\frac{\pi}{12}+\frac{\sqrt{3}}{2}}\right) is a maximum.

\left(\dfrac{5\pi}{6},9e^{\frac{5\pi}{12}-\frac{\sqrt{3}}{2}}\right) is a maximum.

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