Derivatives Of Functions: Some Important Forms

DERIVATIVES OF FUNCTIONS IN PARAMETRIC FORMS

Sometimes the relation between two variables is neither explicit nor implicit, but some link of a third variable with each of the first two variables, separately, establishes a relation between the first two variables. In such a situation, we say that the relation between them is expressed parametrically. The third variable is called the parameter. More precisely, a relation expressed between two variables x and y in the following form x = f (t), y = g(t) is said to be parametric form with 't' as a parameter. In order to find differentiation of functions in such form, we have by chain rule.

$\frac{dy}{dt}=\frac{dy}{dx}.\frac{dx}{dt}$

$or\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\left(whenever\frac{dx}{dt}\ne 0\right)$

\begin{align}Thus\dfrac{dy}{dx}\,=\,\dfrac{g'(t)}{f'(t)}\left(as\dfrac{dy}{dt}\,=\,g'(t)\,and\,\dfrac{dx}{dt}\,=\,f'(t) \right) \\ \left[ provided\,f'(t)\,\ne \,0 \right] \\\end{align}

Example -1: Find $\frac{dy}{dx}$ of x = 2at² , y = 2at³

Solution: $\frac{dx}{dt}=4at$, $\frac{dy}{dt}=6a{t}^2$

$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=\frac{6a{t}^2}{4at}=\frac{3t}{2}$

DERIVATIVES OF A COMPOSITE FUNCTION

Given a composite function y = f(x), i.e. a function represented by y = F(u), u = $\varphi$(x) or y = F [$\varphi$(x)], then $y^{\prime }=\frac{dy}{dx}=\frac{dF}{du}.\frac{du}{dx}$

This is called the chain rule. The rule can be extended to any number of composite variables;

e.g., if y = f(u(v)), then $y^{\prime }=\frac{dy}{dx}=\frac{df}{du}.\frac{du}{dv}.\frac{dv}{dx}$

Example -2: Find dy/dx

(i) $\sin \sqrt{\cos x}$ (ii) $\sin \left(\log \sqrt{\frac{x}{x+1}}\right)$

(iii) $\log \left(x+e^{\sqrt{x}}\right)$

Solution:

(i)

\begin{align}  \dfrac{dy}{dx}=\dfrac{d\left( \sin \sqrt{\cos x}\right)}{d\left( \sqrt{cos x} \right)}.\dfrac{d\left( \sqrt{cos x} \right)}{d\left( \cos x\right)}.\dfrac{d\left( cos x \right)}{dx} \\ \,\,\,\,\,\,\,=\,\,\cos \sqrt{cosx}.\dfrac{1}{2\sqrt{cos x}}.(-\sin x)=-\dfrac{\sin x.cos x\sqrt{cos x}}{2\sqrt{cos x}} \\\end{align}

(ii) Let $\text{y = sin}\left(\text{log}\sqrt{\frac{\text{x}}{\text{x+1}}}\right)$

$\begin{align} & \dfrac{dy}{dx}=\cos \left( \log \sqrt{\dfrac{x}{x+1}} \right)\dfrac{d}{dx}\left( \log \sqrt{\dfrac{x}{x+1}} \right) \\ & \,\,\,\,\,\,\,=\,\,\,\cos \left( \log \sqrt{\dfrac{x}{x+1}} \right)\dfrac{1}{2}\dfrac{d}{dx}\left[ \log x-\log (x+1) \right] \\ & \,\,\,\,\,\,=\,\,\,\dfrac{1}{2}\cos \left( \log \sqrt{\dfrac{x}{x+1}} \right)\left[ \dfrac{1}{x}-\dfrac{1}{x+1} \right] \\ & \,\,\,\,\,\,=\,\,\dfrac{1}{2}\cos \left( \log \sqrt{\dfrac{x}{x+1}} \right)\dfrac{1}{x(x+1)} \\\end{align}$

(iii) $y=\log \left(x+e^{\sqrt{x}}\right)$

$=\frac{1}{x+e^{\sqrt{x}}}\left[1+e^{\sqrt{x}}\frac{d}{dx}\left(\sqrt{x}\right)\right]$=$=\frac{1}{x+e^{\sqrt{x}}}\left[1+e^{\sqrt{x}}\frac{1}{2\sqrt{x}}\right]$

$=\frac{2\sqrt{x}+e^{\sqrt{x}}}{2\sqrt{x}\left(x+e^{\sqrt{X}}\right)}$

DERIVATIVE OF AN IMPLICIT FUNCTION

If x and y are related by the rule F (x, y) = 0 such that y cannot be obtained entirely or exactly in terms of x then y is said to be an implicit function of x.

For example: $x^2+y^2=r^2\Rightarrow y=\pm \sqrt{a^2-x^2}$

Here we do not get a unique value of y for each x.

Eg. x² – y³ + 3x²y = 0

Here also y cannot be obtained entirely in terms of x. To find $\frac{dy}{dx}$ in such cases start differentiating the given equation as it is (using rule of composite functions)

For example: for x³ + y² = a², we have 3x² + 2y$\frac{dy}{dx}$= 0 $\Rightarrow$ $\frac{dy}{dx}=\frac{-3x^2}{2y}$

If y can be expressed entirely in terms of x, then y is said to be an explicit function of x. Note that every explicit function can be written as the implicit function y – f(x) = 0.

Example -3: Find dy/dx

(i) log(xy) = x² + y² (ii) x + y = sin (xy)

Solution: (i) log (xy) = x² + y² $\Rightarrow$ logx + logy = x² + y²

Differentiating w.r.t x

\begin{align}\dfrac{1}{x}+\dfrac{1}{y}\dfrac{dy}{dx}=2x=2y\dfrac{dy}{dx}\Rightarrow \left( \dfrac{1}{y}-2y \right)\dfrac{dy}{dx}=2x-\dfrac{1}{x} \\ \dfrac{1-2{{y}^{2}}}{y}\dfrac{dy}{dx}=\dfrac{2{{x}^{2}}-1}{x}\Rightarrow \dfrac{dy}{dx}=\dfrac{y(2{{x}^{2}}-1)}{x(1-2{{y}^{2}})} \\ \end{align}

(ii) x + y = sin (xy)

Differentiating w.r.t. x, we get 1 + $\frac{dy}{dx}$ = cos (xy).$\left(x\frac{dy}{dx}+y\right)$

\begin{align}  \Rightarrow \,\,\,\,\dfrac{dy}{dx}\left[ 1-x\,\cos (xy) \right]\,=\,y\cos (xy)-1 \\ \Rightarrow \,\,\,\,\dfrac{dy}{dx}=\dfrac{y\,\cos (xy)-1}{1-x\cos (xy)} \\ \end{align}

INVERSE FUNCTIONS AND THEIR DERIVATIVES

Theorem: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if f'(x) exists and f'(x) $\ne$ 0 then g'(y) = $\frac{1}{f^{\prime }(x)}$. This result can also be written as, if $\frac{dy}{dx}$ exists and $\frac{dy}{dx}\ne 0$, then $\frac{dx}{dy}=\frac{1}{\left(\frac{dy}{dx}\right)}$ or $\frac{dy}{dx}$$.\frac{dx}{dy}$=1 or $\frac{dy}{dx}=\frac{1}{\left(\frac{dx}{dy}\right)}\left[\frac{dx}{dy}\ne 0\right]$

Result:

1. $D({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}},-1<x<1$

2. $D({\cos }^{-1}x)=\frac{-1}{\sqrt{1-x^2}},-1<x<1$

3. $D({\tan }^{-1}x)=\frac{1}{1+x^2},x\in R$

4. $D({\sec }^{-1}x)=\frac{1}{\left|x\right|\sqrt{x^2-1}},\left|x\right|>1$

5. $D(\text{cose}{\text{c}}^{-1}x)=\frac{-1}{\left|x\right|\sqrt{x^2-1}},\left|x\right|>1$

6. $D({\cot }^{-1}x)=\frac{-1}{1+x^2},x\in R$

Example -4: Find $\frac{dy}{dx}$ of

(a). y = sin 2x (b). y = cos 2x

(c). y = x sin^–1 x (d). y = a tan^–1 x

(e). y = sec² x (f). y = (log₂ x)²

Solution: (a). $\frac{dy}{dx}$ = 2 cos 2x

(b). $\frac{dy}{dx}$ = – 2 sin 2x

(c). $\frac{dy}{dx}$ = x.$\frac{1}{\sqrt{1-x^2}}$ + sin^–1 x

(d). $\frac{dy}{dx}$ = a$\frac{1}{1+x^2}$

(e). $\frac{dy}{dx}$ = 2 sec x . sec x tan x = 2 sec² x tan x

(f). $\frac{dy}{dx}$ = 2 log₂ x x $\frac{1}{x}$log_e2 = 2log₂x . $\frac{1}{x}$log₂e

HIGHER ORDER DERIVATIVES

Let y = f(x)

First derivative $\frac{d^{}y}{dx}=f^{{}^{\prime }}(x)=y_1$

Second derivative$\frac{d^{2}y}{d{x}^2}=f^{{}^{\prime \prime }}(x)=y_2$

Third derivative$\frac{d^{3}y}{d{x}^3}=f^{{}^{\prime \prime \prime }}(x)=y_3$etc.

Example -5: Find the second derivative of ax³ + bx² + cx + d.

Solution : Let y = ax³ + bx² + cx + d.

$\Rightarrow \frac{dy}{dx}=3a{x}^2+2bx+c$

$\Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=3a(2x)+2b\left(1\right)+0$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=6ax+2b$

Functions of Logarithmic Form

If u and v are functions of independent variable x then to differentiate the functions like u^v , first we take the log and then differentiate.

Example -6: Differentiate ${\left(\log x\right)}^{\tan x}$ w.r.t. sin( m cos^{– 1}x).

Solution: Let y = ${\left(\log x\right)}^{\tan x}$ and z = sin(m cos^{– 1}x)

So logy = tanx log log x

$\frac{1}{y}.\frac{dy}{dx}={\sec }^{2}x.\log \log x+\tan x\times \frac{1}{\log x}\times \frac{1}{x}$

$\frac{dy}{dx}={\left(\log x\right)}^{\tan x}\left({\sec }^{2}x.\log \log x+\frac{\tan x}{x\log x}\right)$

and $\frac{dz}{dx}=\cos \left(m{\cos }^{-1}x\right)\times m\times \frac{-1}{\sqrt{1-x^2}}=\frac{-m\cos \left(m{\cos }^{-1}x\right)}{\sqrt{1-x^2}}$

so $\frac{dy/dx}{dz/dx}=\frac{-{\left(\log x\right)}^{\tan x}\left({\sec }^{2}x.\log \log x+\frac{\tan x}{x\log x}\right)\sqrt{1-x^2}}{m\cos \left(m{\cos }^{-1}x\right)}$.

Differentiating a function w.r.t. to another function

Let we have to differentiate f(x) with respect to g(x). If y = f(x) and t = g(x) then we have to find $\frac{dy}{dx}.$First we find $\frac{dy}{dx}$and $\frac{dt}{dx}$ then $\frac{dy}{dt}=\frac{dy}{dx}÷\frac{dt}{dx}$

Example -7: Differentiate sin²x w. r. t. (logx)²

Solution: Let y = sin²x and t = (logx)²

^{$\frac{dy}{dx}2\sin x\cos x,\frac{dt}{dx}=2\log x\left(\frac{1}{x}\right)=\frac{2\log x}{x}$}

$\frac{dy}{dt}=\frac{dy}{dx}÷\frac{dt}{dx}=\frac{2\sin x\cos x}{2\log x}.x=\frac{x\sin x\cos x}{\log x}$

Example -8: If x² + y² + xy = 2, find $\frac{dy}{dx}.$

Solution: x² + y² + xy = 2,

Differentiating both sides we get,

\begin{align}  \dfrac{d}{dx}\left( {{x}^{2}} \right)+\dfrac{d}{dx}\left( {{y}^{2}} \right)+\dfrac{d}{dx}(xy)=\dfrac{d}{dx}(2) \\ or2x+2y\dfrac{dy}{dx}+\left\{ \dfrac{dx}{dx} \right\}y+x\left\{ \dfrac{d}{dx}y \right\}=0 \\ or2x+2y\dfrac{dy}{dx}+1.y+x.\dfrac{dy}{dx}=0 \\ or(2y+x)\dfrac{dy}{dx}=-(2x+y) \\ or\dfrac{dy}{dx}=-\dfrac{(2x+y)}{(2y+x)} \\\end{align}

Example -9: If 5f(x) + 3f $\left(\frac{1}{x}\right)$ = x + 2 and y = xf (x), then find $\frac{dy}{dx}$ at x = 1.

Solution: Here, $5f(x)+3f\left(\frac{1}{x}\right)=x+2$

Put x = $\frac{1}{x}$ we get,

$5f\left(\frac{1}{x}\right)+3f(x)=\frac{1}{x}+2$

Solving (i) and (ii) we get,

$16f(x)=5x-\frac{3}{x}+4$

$\therefore$ y = x f (x)

$\therefore$ $y=x.\frac{1}{16}\left\{5x-\frac{3}{x}=4\right\}$

$y=\frac{1}{16}\left\{5x^2-3+4\right\}$

or $\frac{dy}{dx}=\frac{1}{16}\left\{10x+4\right\}$

Now $\frac{dy}{dx}$ at x = 1

${\left(\frac{dy}{dx}\right)}_{atx=1}=\frac{10+4}{16}=\frac{14}{16}=\frac{7}{8}$

$\Rightarrow$ ${\left(\frac{dy}{dx}\right)}_{atx=1}=\frac{7}{8}$

Example -10: Find a, b, c and d, where f (x) = (ax + b) cos x + (cx + d) sin x and f' (x) = x cos x is identity in x.

Solution: Here,

f' (x) = x cos x

$\Rightarrow$ a cos x -(ax + b) sin x + c sin x + (cx + d) cos x º x cos x

or (a + cx + d) cos x + (-ax – b + c) sin x º x cos x + 0.sin x

$\Rightarrow$ a + b + cx = x and -ax-b + c = 0

Which is again identity in 'x'

$\Rightarrow$ a + b = 0, c = 1, -a = o, -b + c = o

a =0, b = 1, c = 1, d = 0