Subject: Calculus

# Derivative Examples

Solving mathematical problems especially Calculus regarding derivative and integration sometimes poses difficulty especially for first timers. But according to experts, mathematical solving skills can easily be acquired through frequent exercises and practice solving problems. In these section, let me provide you with more examples that may help you in better understanding of this topic.

## First Order Derivatives

### Example #1

Find the first order derivative of the following function.

f(x)= \frac{x^{2}+bx}{x}

Quotient rule suggests that

 f'(x) = \frac{x \cdot D_{x}(x^{2}+bx)-(x^{2}+bx)\cdot D_{x}(x)}{x^{2}} = \frac{x \cdot (2x+b)-(x^{2}+bx)\cdot (1)}{x^{2}} = \frac{x \cdot (2x+b)-x(x +b) }{x^{2}} = \frac{x [ (2x+b)-(x +b)] }{x^{2}} = \frac{ 2x+b-x +b }{x} = \frac{2x-x}{x} = 1.

That was a lengthy solution, isn't it? But note that had we simplified the function first, the solution would have been very easy. We may even skip the use of quotient rule. Here's how,

 f(x) = \frac{x^{2}+bx}{x} = \frac{x(x+b)}{x} = x+b.

So, differentiating the above simplified form gives

 f'(x) = D_{x}(x+b) = 1.

which is just equal to our initial result.

### Example #2

Find the first order derivative of the following function.

y = \frac{1}{4}x^{4}+\frac{1}{3}x^{3}+\frac{1}{2}x^{2} +x

Applying chain rule directly, we obtain \begin{eqnarray}

 y' = D_{x}( \frac{1}{4}x^{4}+\frac{1}{3}x^{3}+\frac{1}{2}x^{2} +x ) = 4\cdot \frac{1}{4}x^{3}+3\cdot \frac{1}{3}x^{2}+2\cdot \frac{1}{2}x +1 = x^{3}+ x^{2}+x+1.

Thus, the derivative of y is x^{3}+ x^{2}+x+1.

### Example #3

Solve for the first order derivative of the function below.

y=\frac{sin{ax}}{\cos{ax}}-\tan{ax}

Look, one might attempt to differentiate $y$ directly but notice that

 y = \frac{sin{ax}}{\cos{ax}}-\tan{ax} = \tan{ax}-\tan{ax} = 0.

Since the derivative of constant (zero) is zero, then

y'=D_{x}\left ( \frac{sin{ax}}{\cos{ax}}-\tan{ax} \right )=0.

### Example #4

Differentiate the following function once.

g(x) = \cos^{3}(5+x)

We will use chain rule theorem and theorems on derivative of a function such that

 g'(x) = D_{x} \cos^{3}(5+x) = 3\cdot \cos^{2}(5+x)\cdot D_{x}[\cos(5+x)] = -3\cos^{2}(5+x)\sin(5+x).

## Implicit Derivatives

### Example #5

Find the first order derivative of the following implicit function.

x^{2}+4y^{2}-4x-8y+6=0

SOLUTION: We differentiate term by term to get

 2x+8yy'-4-8y' = 0 8yy'-8y' = 4-2x 8y'(y-1) = 4-2x y' = "+\frac{2(2-x)}{8(y-1)}+' y' = \frac{2-x}{y-1}

### Example #6

Differentiate implicitly the following function.

2y^{3}-9y^{2}+24y=3x^{2}-12x

SOLUTION: Applying the same method as shown above, we have

 6y^{2}y'-18yy'+24y' = 6x-12 y'(6y^{2}-18y+24 = 6x-12 y' = \frac{6x-12}{6y^{2}-18y+24}

Thus, the derivative is y'= \frac{6x-12}{6y^{2}-18y+24}

### Example #7

Differentiate implicitly the following function.

x^{2}+y^{2}=6

SOLUTION: Again, repeating the same process, we obtain

 2x+2yy' = 0 2yy' = -2x y' = -\frac{x}{y}.

### Example #8

Differentiate implicitly the following function.

x=\sin(x+y)

This exercise involves trigonometric differentiation which can easily be solved by

 1 = \cos(x+y)\cdot (1+y') \frac{1}{\cos{x+y}} = 1+y' \sec{x+y} = {$1+y'$| \sec(x+y)-1 = y'.

## More Higher Order Derivative Examples

### Example #9

Find y'' when y=\sin{x}+\cos{x}

y=\sin{x}+\cos{x}

and show that y+y''=0 .

SOLUTION:

We first find its first order derivative

 y' = D_{x}(\sin{x}+\cos{x}) = \cos{x}-\sin{x},

then its second order derivative is

 y'' = D_{x}(-\sin{x}+\cos{x}) = -\cos{x}-\sin{x}.

Now, our original equation is y=\sin{x}+\cos{x}. Inserting this into y+y'' along with equation above, we have

 y-y'' = \sin{x}+\cos{x}-\cos{x}-\sin{x} = 0.

### Example #10

Find the third order derivative.

y=\frac{45}{x^2}.

SOLUTION:

Let us first calculate its first derivative.

 y' = D_{x}[45x^{-2}] = -2.\cdot 45x^{-3} y' = -90x^{-3}

Let us then find its second derivative.

 y'' = D_{x}[y'] = -90D_{x}[x^{-3}] = -3\cdot (-90)x^{-4} y'' = 270x^{-4},

Then finally solve for its third derivative which would be,

 y''' = D_{x}[y''] = 270D_{x}[x^{-4}] y''' = 1080x^{-5} = \frac{1080}{x^{-5}}.

Thus, the third derivative is y'''=\frac{1080}{x^{-5}}.

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