simple but the best way of finding derivatives of logarithmic, exponential and inverse trigonometric or circular functions | mathematics made easier.

In this post, we will be following Basic mathematics of grade 11, a book by BC Bajracharya.

Click here to learn basic concepts of derivatives of functions.

Exponential function:

 The function of the form, `f = b^i`, where b refers to base and i refers to index, is called the exponential function. Usually, the base takes a constant value and the index takes variables.

For exmaple,

(1) `y = a^x ...(1)`

where, y = f(x) = function, b = constant value, x = variable of y.

(2) `y = e^x` , this is also known as natural exponential function.

Logarithmic function:

The inverse of an exponential function is called the logarithmic function. If we take log on both sides of the above equation (1) with base 'a' then 

`log_ay = log_a(a^x)`

or, `log_ay = x log_aa`

[`\therefore` `log_aa^b = b log_aa` and `log_a a = 1`.]

So, `log_ay = x .....(2)`

In summary, if `y = a^x` is an exponential function, then `log_ay = x` is a logarithmic function.

In most cases, a special type of exponential function `y = e^x` is used. Here,  e is given as the limiting value as below:

`e = \lim_(n->\infty)(1 + \frac{1}{n})^n ........(3)`

Numerically, `n \approx 2.718281828`. 

Logarithmic function with base e or `log_e` is also called natural log function and is denoted by ln or simply log.

Some properties of conversion of logarithmic functions:

(1) `log_a(x\cdoty) = log_ax + log_ay`

(2) `log_a(\frac{x}{y}) = log_ax - log_ay`

(3) `log_a(x^n) = n log_ax`

(3) `log_aa = 1`

(4) `log_a1 = 0`

(5) ` log_ap = log_aq \cdot log_qp`

Limit Theorems of logarithmic and exponential functions:

1) `\lim_(x->0)\frac{log(1+x)}{x} = 1`

2) `\lim_(x->0)\frac{e^x - 1}{x} = 1`

3) `\lim_(x->0)\frac{a^x - 1}{x} = 1` 

You can simply prove these formulae by using L-Hospital's rule.

Derivatives of some logarithmic functions:

1.  Derivative of  `y = logx`.

Let's `\Deltax`  and `\Deltay` be small increments or changes in x and y respectively. Then

`y = logx ....(1)` and

`y + \Deltay = log(x + \Deltax) .......(2)`

Subtrac (1) from (2), then

`\Deltay = \frac{log(x+\Deltax)-logx}{\Deltax}`

or, `\frac{\Deltay}{\Deltax} = \frac{log(\frac{x+\Deltax}{x})}{\Deltax}`

[`\therefore` Use of property (2).]

`\frac{\Deltay}{\Deltax} = \frac{log(1+\frac{\Deltax}{x})}{\Deltax}`

  Taking limit as `\Deltax->0` on both sides,

`\lim_(\Deltax->0)\frac{\Deltay}{\Deltax} = \lim_(\Deltax->0)\frac{log(1+\frac{\Deltax}{x})}{\Deltax}`

or,  `\frac{dy}{dx} = \lim_(\Deltax->0) \frac{log(1+\frac{\Deltax}{x})}{\frac{\Deltax}{x} \cdot x}`

Using the limit theorem, `\lim_(x->0)\frac{log(1+x)}{x} =1`, 

or, `\frac{dy}{dx}= 1 \cdot \frac{1}{x}`

`\therefore` `\frac{d}{dx} logx = \frac{1}{x}`.

2. Derivative of `y = log(ax+b)`. 

To get the derivative of the given function, let's use rules of derivatives as well as the limit theorem as used in the above problem.

Let's `y = log(ax+b) ....(1)` 

Differentiating (1) with respect to x, 

`\frac{d}{dx}y = \frac{d}{dx} log(ax+b)`

Here we use chain rule as, learn rules from here.

`\frac{d}{dx}y = \frac{d log(ax+b)}{d(ax+b)} \cdot \frac{d (ax+b)}{dx}`

Using `\frac{d logx}{dx} = \frac{1}{x}`,

`\frac{d y}{dx} = \frac{1}{(ax+b)} \cdot (a \cdot 1 + 0)`

or, `\frac{d y}{dx} = \frac{a}{ax+b}`

`\therefore` `\frac{d log(ax+b)}{dx} = \frac{a}{ax+b}`.

By using limit theorem,

Here, y = log(ax+b) .... (1)

Also, `y + \Deltay = log(a(x+\Deltax)+b) .....(2)`

Subtract (1) from (2),

`\Deltay = log(a(x+\Deltax)+b) - log(ax+b`

or, `\frac{\Deltay}{\Deltax} = \frac{log(a(x+\Deltax)+b) - log(ax+b)}{\Deltax}`

or, `\frac{\Deltay}{\Deltax} = \frac{log(\frac{a(x+\Deltax)+b}{ax+b})}{\Deltax}`

or, `\frac{\Deltay}{\Deltax} = \frac{ log(1+\frac{a\Deltax}{ax+b})}{\Deltax}`

Now taking limit as `\Deltax->0` on both sides,

`\lim_(\Deltax->0)\frac{\Deltay}{\Deltax} = \lim_(\Deltax->0)\frac{log(1+\frac{a\Deltax}{ax+b})}{\Deltax}`

or, `\frac{dy}{dx} = \lim_(\Deltax->0)\frac{log(1+\frac{a\Deltax}{ax+b})}{\frac{a\Deltax}{ax+b}} \frac{a}{ax+b}`

or, `\frac{dy}{dx} = 1 \cdot \frac{a}{ax+b}`

`\therefore` `\frac{d}{dx} log(ax+b) = \frac{a}{ax+b}`.

3. Derivative of `y = log(\frac{x}{10})`

Here, `y = log(\frac{x}{10}) ......(1)`

Also, `y+\Deltay = log(\frac{x+\Deltax}{10}) ....(2)`

Subtract (1) from (2),

`\Deltay = log(\frac{x+\Deltax}{10}) - log(\frac{x}{10})`

or, `\Deltay = log[\frac{\frac{x+\Deltax}{10}}{\frac{x}{10}}]`

Here upper 10 cancels lower 10 insid log, so

`\Deltay = log[\frac{x+\Deltax}{x}]`

or. `\frac{\Deltay}{\Deltax} = \frac{log[\frac{x+\Deltax}{x}]}{\Deltax}`

or, `\frac{\Deltay}{\Deltax} = \frac{log[1+\frac{\Deltax}{x}]}{\Deltax}`

Now taking limit as `\Deltax->0` on both sides,

`\lim_(\Deltax->0)\frac{\Deltay}{\Deltax} = \lim_(\Deltax->0) \frac{log[1+\frac{\Deltax}{x}]}{\frac{\Deltax}{x}} \frac{1}{x}`    

or, `\frac{dy}{dx} = 1 \cdot \frac{1}{x}`

`\therefore` `\frac{d}{dx} log(\frac{x}{10}) = \frac{1}{x}.`

Derivatives of some Exponential functions:

1. Derivative of  `y = e^x`

Here, `y = e^x ....(1)`

Also, `y+\Deltay = e^(x+\Deltax) ....(2)`

Subtract (1) from (2),

`\Deltay = e^(x+\Deltax) - e^x` 

or, `\Deltay = e^x (e^(\Deltax) - 1)`

or, `\frac{\Deltay}{\Deltax} = \frac{e^x (e^(\Deltax) - 1)}{\Deltax}`

or, `\lim_(\Deltax->0) \frac{|Deltay}{\Deltax} =e^x  \lim_(\Deltax->0) \frac{e^(\Deltax) - 1}{\Deltax}`

Using limit theorem, `\lim_(x->0)\frac{e^x - 1}{x} =1`,

or, `\frac{dy}{dx} = e^x \cdot 1`

`\therefore` `\frac{d}{dx} e^x = e^x`.

2. Derivative of  `y = e^(ax+b)`

Here, `y = e^(ax+b) ..... (1)`

Also, `y+\Deltay = e^[a(x+\Deltax)+b] ....(2)`

Subtract (1) from (2),

`\Deltay = e^[a(x+\Deltax)+b] - e^[ax+b]`

or, `\Deltay= e^[ax+b] \cdot e^[a\Deltax] - e^[ax+b]`

or, `\Deltay = e^[ax+b] (e^[a\Deltax] - 1)`

or, `\frac{\Deltay}{\Deltax} = e^[ax+b] \frac{e^[a\Deltax] - 1}{\Deltax}`

or, `\lim_(\Deltax->0) \frac{\Deltay}{\Deltax} = e^[ax+b] \lim_(\Deltax->0) \frac{e^[a\Deltax] - 1}{a\Deltax} a`

or, `\frac{dy}{dx} = e^[ax+b] \cdot a`

`\therefore` `\frac{d}{dx} e^[ax+b] = a e^[ax+b]`.

3. Derivative of  `y = e^[\frac{x}{3}]`

Here, `y = e^[\frac{x}{3}] ....(1)`

Also, `y+\Deltay = e^[\frac{x+\Deltax}{3}] .......(2)`

Subtract (1) from(2),

`\Deltay = e^[\frac{x+\Deltax}{3}] - e^[\frac{x}{3}]`

or, `\Deltay = e^[\frac{x}{3}] (e^[\frac{\Deltax}{3}] - 1)`

or, `\frac{\Deltay}{\Deltax} = e^[\frac{x}{3}] \frac{e^[\frac{x}{3}] - 1}{\Deltax}`

or, `\lim_(\Deltax->0)\frac{\Deltay}{\Deltax} = e^[\frac{x}{3}] \lim_(\Deltax->0) \frac{e^[\frac{\Deltax}{3}] -1}{\frac{\Deltax}{3}} \times \frac{1}{3}`

or, `\frac{dy}{dx} = e^[\frac{x}{3}] \times 1 \times \frac{1}{3}`

`\therefore` `\frac{d}{dx} e^[\frac{x}{3}] = \frac{1}{3} e^[\frac{x}{3}]`.

Inverse trigonometric functions or inverse circular functions:-

As you all knew before, the basic trigonometric functions are

y = sin(x), y = cos(x) and y = tan(x).

If these functions are expressed as

siny = x, cosy = x and tany = x so that,

`y = sin^-1x`, `y = cos^-1x` and `y = tan^-1x`, read as inverse sine of x, inverse cosine of x and inverse tangent of x respectively. These functions are called inverse trigonometric or inverse circular functions.

These inverse circular functions are also denoted by using arcs such as,

y = arc sinx = `sin^-1x` and same for cosine, tangent etc.

Note that: `sin^-1x` means never take in mind as `\frac{1}{sinx}` because it is totally wrong. `sin^-1x` and `\frac{1}{sinx}` are not the same.

Some useful relations of inverse circular funations:

1.  `sin^-1sin\theta = \theta` and similar for other  functions
2. `sin^-1x = cosec^-1(\frac{1}{x})` and same for other functions
3. `sin^-1x + cos^-1x = \frac{\pi}{2}`
4. `sin^-1x = cos^-1\sqrt{1-x^2} = tan^-1\frac{x}{\sqrt{1-x^2}}`
5. `sin^-1(-x) = - sin^-1x` and `cos^-1(-x) = \pi - cos^-1x`

Derivatives of inverse circular functions:

1. Derivative of `y=sin^-1x`

Here, `y = sin^-1x ......(1)`

or, `siny = x  ....(2)`

Differentiating (2) both sides w.r.t y,

`\frac{d siny}{dy} = \frac{dx}{dy}`

or, `cosy = \frac{dx}{dy}`

or, `\frac{1}{cosy} = \frac{dy}{dx}`

or, `\frac{dy}{dx} = \frac{1}{cosy}` 

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1 -sin^2y}}`

Using (2),

`\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}`

`\therefore` `\frac{d}{dx} sin^-1x = \frac{1}{\sqrt{1-x^2}}`.

2. Derivative of `y = cos^-1x`.

Here, `y = cos^-1x`

`cosy = x   .......(1)`

`\frac{d cosy}{dy} = \frac{dx}{dy}`

or, `-siny = \frac{dx}{dy}` 

`\frac{d y}{dx} = \frac{-1}{\sqrt{1-cos^2y}}`

`\therefore` `\frac{d}{dx} cos^-1x = \frac{-1}{\sqrt{1 - x^2}}`.

3. Derivative of `y = tan^-1x`.

Here, `y = tan^-1x`

`\therefore` tany = x ...(1)

Differentiating (1) w.r.t y,

`\frac{d}{dy} tany = \frac{dx}{dy}`

or, `sec^2y = \frac{dx}{dy}`

or, `\frac{dy}{dx} = \frac{1}{sec^2y}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1+tan^2y}}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1+x^2}}`

`\therefore` `\frac{d}{dx} tan^-1x = \frac{1}{\sqrt{1+x^2}}`.

4. Derivative of `y = cot^-1x`.

Here, `y = cot^-1x`

`\therefore` coty = x ....(10)

Just follow the steps as in `tan^-1x`, then you get 

`\frac{d}{dx} cot^-1x = \frac{-1}{\sqrt{1+x^2}}`

5. Derivative of `y = sec^-1x`.

Here, `y = sec^-1x`

or, `secy = x  .......(1)`

or, `\frac{d}{dy} secy = \frac{dx}{dy}`

or, `secy tany = \frac{dx}{dy}`

or, `\frac{dy}{dx} = \frac{1}{secy tany}`

or, `\frac{dy}{dx} = \frac{1}{secy \sqrt{sec^2y - 1}}`

or, `\frac{dy}{dx} = \frac{1}{x \sqrt{x^2 - 1}}`

`\therefore` `\frac{dy}{dx} = \frac{1}{x \sqrt{x^2 - 1}}`.

6. Derivative of `y = cosec^-1x`.

Here, `y = cosec^-1x`

or, cosecy = x  .....(1)

or, `\frac{d}{dy} cosecy = \frac{dx}{dy}`

or, `- cosecy coty = \frac{dx}{dy}`

or, `\frac{dy}{dx} = \frac{-1}{cosecy coty}`

or, `\frac{dy}{dx} = \frac{-1}{cosecy  \sqrt{cosec^2y -1}}`

or, `\frac{dy}{dx} = \frac{-1}{x \sqrt{x^2 - 1}}`

`\therfore` `\frac{d}{dx} cosec^-1x = \frac{-1}{x \sqrt{x^2 - 1}}`.

7. Find the derivative of `y = sin^-1(1-2x^2)`.

Here, `y = sin^-1(1-2x^2)`.

or, `siny = 1-2x^2`

or, `\frac{d}{dy} siny = \frac{d}{dy}(1-2x^2)`

or, `cosy = -2 \times 2x \frac{dx}{dy}`

or, `\frac{dy}{dx} = \frac{-4x}{cosy}`

or, `\frac{dy}{dx} = \frac{-4x}{\sqrt{1-sin^2y}}`

or, `\frac{dy}{dx} = \frac{-4x}{\sqrt{1 - (1-2x^2)^2}}`

or, `\frac{dy}{dx}  = \frac{-4x}{\sqrt{1 - 1 + 4x^2 - 4x^4}}` 

or, `\frac{dy}{dx} = \frac{-4x}{\sqrt{4x^2 - 4x^4}}`

or, `\frac{dy}{dx} = \frac{-4x}{2x \sqrt{1-x^2}}`

`\therefore` `\frac{d}{dx} sin^-1(1-2x^2) = \frac{-2}{\sqrt{1-x^2}}`.

8. Find `\frac{dy}{dx}` when `y = cos^-1(\frac{1-x^2}{1+x^2})`.

Here, `y = cos^-1(\frac{1-x^2}{1+x^2})`

or, ` cosy = \frac{1-x^2}{1+x^2} ......(1)`

or, `\frac{d}{dy} cosy = \frac{d}{dy}(\frac{1-x^2}{1+x^2})`

Here you need to use quotient rule of derivative on right hand side.

or, `-siny = \frac{(1+x^2)\frac{d}{dy}(1-x^2) - (1-x^2) \frac{d}{dy}(1+x^2)}{(1+x^2)^2}`

or, `-siny = \frac{(1+x^2)(-2x \frac{dx}{dy}) - (1-x^2)(2x \frac{dx}{dy})}{(1+x^2)^2}`

or, `-siny = \frac{dx}{dy} \frac{-2x(1-x^2) - 2x(1+x^2)}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{1}{-siny} \frac{[-2x(1-x^2 +1 + x^2)]}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1-cos^2y}} \frac{-2x \times 2}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1 - [\frac{1-x^2}{1+x^2}]^2}} \frac{-4x}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{1}{\frac{\sqrt{(1+x^2)^2 - (1-x^2)^2}}{(1+x^2)}} \frac{-4x}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{1+2x^2+x^4 - 1 + 2x^2 - x^4}} \frac{-4x}{1+x^2}`

or, `\frac{dy}{dx} = \frac{1}{\sqrt{4x^2}} (\frac{-4x}{1+x^2})`

or, `\frac{dy}{dx} = \frac{1}{2x} (\frac{-4x}{1+x^2})`

or, `\frac{dy}{dx} = \frac{-2}{1+x^2}`

`\therefore` `\frac{d}{dx} cos^-1(\frac{1-x^2}{1+x^2}) = \frac{-2}{1+x^2}`.

9. What is value of `\frac{dy}{dx}` if  `y = tan^-1(\frac{sin2x}{1+cos2x}) ?`

Here, `y = tan^-1(\frac{sin2x}{1+cos2x})`

So, `tany = \frac{sin2x}{1+cos2x}   ......(1)`

`\frac{d}{dy} tany = \frac{d}{dy} \frac{sin2x}{1+cos2x}`

or, `sec^2y = \frac{(1+cos2x) 2cos2x \frac{dx}{dy} - sin2x (-2sin2x) \frac{dx}{dy}}{(1+cos2x)^2}`

or, `sec^2y = \frac{\frac{dx}{dy}[2cos2x + 2cos^2(2x) + 2 sin^2(2x)]}{(1+cos2x)^2}`

or, `\frac{dy}{dx} = \frac{2cos2x +2(cos^2(2x)+sin^2(2x))}{sec^2y \times (1+cos2x)^2}`

or, `\frac{dy}{dx} = \frac{2cos2x + 2}{(1+tan^2y)\times(1+cos2x)^2}`

or, `\frac{dy}{dx} = \frac{2(cos2x+1)}{(1+(\frac{sin2x}{1+cos2x})^2)(1+cos2x)^2}`

or, `\frac{dy}{dx} = \frac{2(1+cos2x)}{\frac{(1+cos2x)^2 + sin^2(2x)}{(1+cos2x)^2}\times (1+cos2x)^2}`

or, `\frac{dy}{dx} = \frac{2(1+cos2x)}{1 + 2cos2x + (cos^2(2x) + sin^2(2x))}`

or, `\frac{dy}{dx} = \frac{2(1+cos2x)}{1+2cos2x +1}`

or, `\frac{dy}{dx} = \frac{2(1+cos2x)}{2(1+cos2x)} = 1`

`\therefore` `\frac{d}{dx} tan^-1(\frac{sin2x}{1+cos2x}) = 1`.

10. If `tan^-1(\frac{2x}{1-x^2})` is the given function, then find it's derivative.

Here, `y = tan^-1(\frac{2x}{1-x^2})`

or, ` tany = \frac{2x}{1-x^2}  ......(1)` 

or, `\frac{d}{dy}tany = \frac{d}{dy}(\frac{2x}{1-x^2})`

or, `sec^2y = \frac{(1-x^2) \frac{d}{dy}(2x) - (2x) \frac{d}{dy}(1-x^2)}{(1-x^2)^2}`

or, `sec^2y = \frac{(1-x^2)2 \frac{dx}{dy} - 2x (-2x) \frac{dx}{dy}}{(1-x^2)^2}`

or, `sec^2y = \frac{\frac{dx}{dy}\times[2(1-x^2) + 4x^2]}{(1-x^2)^2}`

or, `\frac{dy}{dx} = \frac{2 - 2x^2 +4x^2}{sec^2y \times (1-x^2)^2}`

or, `\frac{dy}{dx} = \frac{2+2x^2}{(1+tan^2y) \times (1-x^2)^2}`

or, `\frac{dy}{dx} = \frac{2(1+x^2)}{[1+(\frac{2x}{1-x^2})^2] \times (1-x^2)^2}`

or, `\frac{dy}{dx}= \frac{2(1+x^2)}{\frac{(1-x^2)^2 +4x^2}{(1-x^2)^2} \times (1-x^2)^2}`

or, `\frac{dy}{dx} = \frac{2(1+x^2)}{(1-x^2)^2 +4x^2}`

or, `\frac{dy}{dx} = \frac{2(1+x^2)}{1-2x^2+x^4+4x^2}`

or, `\frac{dy}{dx} = \frac{2(1+x^2)}{1+2x^2+x^4}`

or, `\frac{dy}{dx} = \frac{2(1+x^2)}{(1+x^2)^2}`

or, `\frac{dy}{dx} = \frac{2}{1+x^2}`

`\therefore` `\frac{d}{dx}tan^-1(\frac{2x}{1-x^2}) = \frac{2}{1+x^2}`.