derivatives of some difficult functions | differentiating complex functions

 If y = f(x) is a function of a variable x, and `\Deltax` and `\Deltay` are small increment or change in x and y respectively, then the derivative of a function f(x) is defined as the ratio of change in function with respect to it's variable x when `\Deltax ->0`. This means the ratio of change in f(x) for a small change in x is called the derivative or differentiation coefficient of f(x) with respect to x.

This post may help those students/learners who are related to the field of mathematics and find derivatives hard to understand.

Let's find the derivatives of some of the difficult functions.

1. Find `\frac{dy}{dx}` when `y = tan^-1(\frac{sinx+cosx}{cosx-sinx})`.

Solution:-

Here, `y = tan^-1(\frac{sinx+cosx}{cosx-sinx}) .... (1)` 

`\frac{dy}{dx} = ?`

Let's simplify (1), take cosx common inside () bracket.

`y = tan^-1(\frac{cosx(\frac{sinx}{cosx}+1)}{cosx(1-\frac{sinx}{cosx})})`

But `\frac{sinx}{cosx} = tanx`, so 

`y = tan^-1(\frac{tanx+1}{1-tanx}) ..... (2)`

We have `tan(\frac{\pi}{4}+x) = \frac{tan\frac{\pi}{4} +tanx}{1-tan\frac{\pi}{4} tanx}`

But, `tan\frac{\pi}{4} =1` so we get

`tan(\frac{\pi}{4}+x) = \frac{1+tanx}{1-tanx}`

Using these results in (2),

`y = tan^-1(tan(\frac{\pi}{4}+x))`

But `tan^-1(tan(\frac{\pi}{4}+x)) = \frac{\pi}{4}+x`, so

`y = \frac{\pi}{4} + x ..... (3)` which is equivalent to (1).

Now let's differentiate (3) w.r.t 'x' then,

`\frac{dy}{dx} \frac{d}{dx}(\frac{\pi}{4}+x)`

or, `\frac{dy}{dx} = \frac{d}{dx}\frac{\pi}{4}+\frac{d}{dx}x`

As you know derivative of any contants or numbers is zero, then

`\frac{d}{dx}\frac{\pi}{4} = 0` which gives us

`\frac{dy}{dx} = 0 +1 =1`

Thus, `\frac{dy}{dx} = 1` //

2. If `y = tan^-1(\frac{\sqrt{1+x^2}-1}{x})`, then `\frac{dy}{dx}` equals ?

Solution:-

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

Let's take, `x = tan\theta` so that 

`\sqrt{1+x^2}-1`

= `\sqrt{1+tan^2\theta}-1`

= `sec\theta -1 ..... (2) ` [`\therefore` `1+tan^2\theta = sec\theta`]

Using this result in (1),

`y = tan^-1(\frac{sec\theta -1}{tan\theta})`

or, `y = tan^-1(\frac{\frac{1}{cos\theta}-1}{\frac{sin\theta}{cos\theta}})`

or, `y = tan^-1(\frac{1-cos\theta}{cos\theta \frac{sin\theta}{cos\theta}})`

or, `y = tan^-1(\frac{1-cos\theta}{sin\theta})`

or, `y = tan^-1(\frac{2 sin^2(\frac{\theta}{2})}{2 sin(\frac{\theta}{2}) cos(\frac{\theta}{2})})`

or, `y = tan^-1(\frac{sin(\frac{\theta}{2})}{cos(\frac{\theta}{2})})`

or, `y = tan^-1(tan(\frac{\theta}{2}))`

or, `y = \frac{\theta}{2} ....(3)`

But as we assumed, `x = tan(\theta)`

`\therefore` `\frac{\theta}{2} = \frac{1}{2} tan^-1(x)`

Using this in (3),

`y = \frac{1}{2} tan^-1(x) ....(4)`

Now differentiating (4) 2.r.t 'x', 

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

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

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

3. Find `\frac{dy}{dx}` at x = e when `y = \sqrt{x log_e(x)}`. 

Solution:-

Here,  `y = \sqrt{x log_e(x)}`

or, `y = (x log_e(x))^\frac{1}{2} ...(1)`

Take log of (1) with `log_e = log`,

`logy = \frac{1}{2} [logx +  log(logx)]`

Takiing w.r.t 'x' on both sides,

`\frac{d}{dx}logy = \frac{1}{2} \frac{d}{dx}[logx + log(logx)]`

or, `\frac{1}{y} \frac{dy}{dx} = \frac{1}{2}[\frac{1}{x} + \frac{1}{logx} \cdot \frac{1}{x}]`

or, `\frac{dy}{dx} = y \frac{1}{2} [\frac{1}{x} + \frac{1}{x logx}]`

At x = e, 

`\frac{dy}{dx}|_(x=e) = \sqrt{e log_ee} \frac{1}{2}[\frac{1}{e}+\frac{1}{e loge}]`

or, `\frac{dy}{dx}|_(x=e) = \sqrt{e loge} \frac{1}{2}[\frac{1}{e}+\frac{1}{e loge}]`

But `loge = log_ee = 1`, so

`\frac{dy}{dx}|_(x=e) = \frac{1}{2} \sqrt{e} [\frac{1}{e}+ \frac{1}{e}]`

or, `\frac{dy}{dx}|_(x=e) = \frac{1}{2} \sqrt{e} \frac{2}{e}`

or, `\frac{dy}{dx}|_(x=e) = \frac{1}{\sqrt{e}}`

Thus, `\frac{dy}{dx} = \frac{1}{\sqrt{e}}` at x = e. // 

4. What is the differentiation coefficient of `y = e^(x^3)` ?

Solution:-

Here, `y =e^(x^3)`

Taking log_e on both sidees,

`logy = x^3 loge`

or, `logy = x^3`

Differentiate with respect to x,

`\frac{d}{dx}logy = \frac{d}{dx}x^3`

or, `\frac{1}{y} \frac{dy}{dx} = 3x^2`

or, `\frac{dy}{dx} = 3yx^2`

or, `\frac{dy}{dx} = 3e^(x^3)x^2`

`\therefore` `\frac{dy}{dx} = 3 e^(x^3) x^2` which is reqired differential coefficient. //

5. If `f(x) =\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+ ....... \infty}}}}}`, then what is the value of `\frac{d}{dx}f(x)` ?

Solution:-

Here, `f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+ ...... \infty}}}} .....(1)`

In (1), the series goes to infinite terms. So, the series after 'x+' in the first square root, can be denoted by f(x), that is,

`f(x) \approx \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+ ..... \infty}}}} .....(2)`

Using (2) in (1),

`f(x) = \sqrt{x+f(x)}`

or, `f^2(x) = (x+f(x))`

Differentiate with respect to x,

`\frac{d}{dx}f^2(x) = \frac{d}{dx}(x+f(x))`

or, `2f(x) \frac{d f(x)}{dx} = (1 + \frac{d f(x)}{dx}`

or, `2 f(x) \frac{d f(x)}{dx} - \frac{d f(x)}{dx} = 1`

or, `\frac{d f(x)}{dx}(2f(x) - 1) = 1`

or, `\frac{d f(x)}{dx} = \frac{1}{2f(x) - 1}`

`\therefore` `\frac{d f(x)}{dx} = \frac{1}{(2 \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+ ....... \infty}}}}} - 1)}` //

6. If `y = (sinx)^(tanx)`, then find `\frac{dy}{dx}`.

Solution:-

Here, `y = (sinx)^(tanx) .......(1)`

Taking log on both sides of (1),

`logy = tanx log(sinx) ...(2)` 

Differentiating (2) with respect to x,

`\frac{d}{dx}logy = \frac{d}{dx} tanx log(sinx)`

or, `\frac{1}{y} \frac{dy}{dx} = tanx \frac{1}{sinx} cosx + sec^2x log(sinx)`

or, `\frac{dy}{dx} = y[tanx cotx + sec^2x log(sinx)]`

But tanx cotx = 1,

`\frac{dy}{dx} = y[1 + sec^2x log(sinx)]`

`\therefore` `\frac{dy}{dx} = (sinx)^tanx [1 + sec^x log(sinx)]` // 

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