# 12th Class Mathematics Applications of Derivatives Notes - Mathematics Olympiads -Application of Derivatives

Notes - Mathematics Olympiads -Application of Derivatives

Category : 12th Class

Application of Derivatives

• Continuity and Differentiability of a function: Let a function $y=f(x)$ is said to be continuous at

$x=a$ then $\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\,\,f(x)=f(a)$

Generally, a function is said to be continuous at $x=a$ when the graph of that function can be drawn/sketched without lefting the pencil.

• Differentiation: The process of finding out the differentiability/derivatives of the function $y=f(x)$in the interval (a, b) is said to be differentiation.

• Derivatives of f(x): Let $y=f(x)$ is continuous in interval [a, b]. Let a point $c\in (a,b)$

Then function $y=f(x)$ is differentiable at $x=c$

i.e. $\underset{x\to c}{\mathop{\lim }}\,\frac{f(x+c)-(c)}{c}=f'(c)$

Solved Example

1. Find derivative of $y=\sin x.$ by 1st principle:

Let $y=f(x)=\sin x$    ...(1)

Let $\delta$x be the small increment in x then $\delta$y be the corresponding increment in y.

$y+\delta y=\sin (x+\delta x)$   ...(2)

Now, on subtracting equation (1) from (2), we get

$y+\delta y-y=\sin (x+\delta x)-\sin x$

$\delta y=2.\cos \frac{x+\delta x+x}{2}.\sin \left( \frac{x+\delta x+x}{2} \right)$

Dividing $\delta$x on both sides and taking limit $\delta x\to 0,$we get

$\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{\delta y}{\delta x}=\underset{\delta x\to 0}{\mathop{\lim }}\,2.\frac{\cos \left( \frac{2x+\delta x}{2} \right).\sin \left( \frac{\delta x}{2} \right)}{\delta x}$

$\frac{dy}{dx}\underset{\delta x\to 0}{\mathop{\lim }}\,2.\cos \left( x+\frac{\delta x}{2} \right).\left( \frac{\sin \frac{\delta x}{2}}{\frac{\delta x}{2}\times 2} \right)$

On applying limit $\delta x\to 0,$we get

$\frac{dy}{dx}=\cos x\times 1=\cos x\left[ \underset{\theta \to 0}{\mathop{\lim }}\,\frac{\sin \theta }{\theta }=1 \right]$

Thus,$\frac{d}{dx}(\sin x)=\cos x$

$\frac{dy}{dx}$is said to be differential coefficient of $y=f(x).$ It is denoted by${{y}_{1}}$or f $'(x)$

• Some Important Formulae
1. $\frac{d}{dx}(\sin x)=\cos x$
2. $\frac{d}{dx}(\cos x)=-\sin x$
3. $\frac{d}{dx}(\tan x)={{\sec }^{2}}x$
4. $\frac{d}{dx}(\cot x)=-\cos e{{c}^{2}}x$
5. $\frac{d}{dx}(\sec x)=\sec x\tan x$
6. $\frac{d}{dx}(\cos ec\,x)=-\cos ec\,x.\cot x$
7. $\frac{d}{dx}({{x}^{n}})=n{{x}^{n-1}}$
8. $\frac{d}{dx}({{e}^{x}})={{e}^{x}}$
9. $\frac{d}{dx}(\log x)=\frac{1}{x}$
10. $\frac{d}{dx}(si{{n}^{-1}}x)=\frac{1}{\sqrt{1-{{x}^{2}}}}$
11. $\frac{d}{dx}(co{{s}^{-1}}x)=\frac{-1}{\sqrt{1-{{x}^{2}}}}$
12. $\frac{d}{dx}(ta{{n}^{-1}}x)=\frac{1}{1+{{x}^{2}}}$
13. $\frac{d}{dx}(x)=1$
14. $\frac{d}{dx}(C)=0$ where $C=$any const.
15. $\frac{d}{dx}(\sin ax)=a\,\text{cos}\,ax$
16. $\frac{d}{dx}({{a}^{x}})={{a}^{x}}.\log a$

• Some Basic Rules of Differentiation
1. $\frac{d}{dx}(u\,\pm v)=\frac{d}{dx}(u)\pm \frac{d}{dx}(v)$where u and v be the function of x.
2. $\frac{d}{dx}(C.u)=C.\frac{d}{dx}(u)$ where $C=$any const.

e.g. $\frac{d}{dx}(5{{x}^{2}})=5.\frac{d}{dx}({{x}^{2}})=5[2.{{(x)}^{2-1}}]=10{{x}^{1}}=10x$

1. $\frac{d}{dx}(u.v.)=u.\frac{d\text{v}}{dx}+\frac{du}{dx}$

e.g. $\frac{d}{dx}({{e}^{x}}.\sin x)={{e}^{x}}.\frac{d}{dx}(\sin x)+\sin x\frac{d}{dx}({{e}^{x}})={{e}^{x}}\cos x+\sin x.{{e}^{x}}$

$={{e}^{x}}(\cos x+\sin x)$

1. $\frac{d}{dx}\left( \frac{u}{\text{v}} \right)=\frac{\text{v}.\frac{du}{dx}-u.\frac{d\text{v}}{dx}}{{{\text{v}}^{2}}}$ e.g. $\frac{d}{dx}\left( \frac{{{x}^{2}}}{\sin x} \right)_{\text{v}}^{u}=\frac{\sin x\frac{d}{dx}({{x}^{2}})-{{x}^{2}}\frac{d}{dx}(\sin x)}{{{\sin }^{2}}x}$
2. $\frac{d}{dx}(lo{{g}_{c}}{{a}^{x}})=\frac{d}{dx({{a}^{x}})}(\log {{a}^{x}}).\frac{d}{dx}({{\underline{a}}^{x}})=\frac{1}{{{a}^{x}}}.{{a}^{x}}.\log a=\log a$

• Tangent & Normals:

Geometrical meaning of derivative at point: The derivative of a function $f(x)$ at a point $x=a$ is the of the tangent of the curve $y=f(x)$ at the point $(a,f(a)).$

Let us consider a curve $y=f(x)$ & take a point P$(x,y)$ on it. We draw the tangent to the curve at $P(x,y)$which makes an angle $\alpha$ with positive direction of x-axis. Then,

${{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,P(x,y)\,=\,tan\,\,\alpha \,=\,m(say)}}$

It is said to be gradient or slope of the tangent to the curve $y=f(x)$at$p(x,y)$.

• Equation of the tangent: The equation of the tangent to a curve $y=f(x)$ at the given point $P({{x}_{1}},{{y}_{1}})$ is written in point slope form of the equation of straight line.

$y-{{y}_{1}}=m(x-{{x}_{1}})$

where m = slope of the tangent

$y-{{y}_{1}}={{\left. \frac{-dy}{dx} \right|}_{\,\,at\,\,P({{x}_{1}},\,{{y}_{1}})\,\,\times \,\,(x-{{x}_{1}}).}}$

• Equation of the Normal: The equation of the normal to the curve$y=f(x)$ at the given point $({{x}_{1}},{{y}_{1}})$ is written as

$y-{{y}_{1}}=\frac{-1}{m}({{x}_{1}}-{{x}_{1}})\,\,y-{{y}_{1}}=\frac{-1}{{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,P({{x}_{1}},{{y}_{1}})}}}\times (x-{{x}_{1}})$

Note: Since tangent & normal to a curve $y=f(x)$ at point $({{x}_{1}},{{y}_{1}})$ is perpendicular to each other at that point.

$\because \,\,\,{{m}_{1}}.{{m}_{2}}=-1$

Where, ${{m}_{1}}=$ slope of the tangent to the curve

${{m}_{2}}=$ slope of the normal to the curve

Solved Examples

1. Find the equation of the tangent and normal to the curve $y=2{{x}^{2}}+3\sin x,$at $x=0.$

Sol. The given equation of the curve is

$y=2{{x}^{2}}+3\sin x$                               ....(1)

Putting$x=0$

then $y=2\times 0+3\text{ }sin0=0.$

Here, the point of the contact is (0, 0).

$\because$ Differentiating (1) w.r.t. x, we get

$\frac{dy}{dx}=2(2x)+3.\cos x{{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,(0,0)}}=4\times 0+3cos({{0}^{\underline{o}}})=0+3=3$

i.e. ${{m}_{1}}=3$

$\therefore$ Equation of the tangent is

$y-{{y}_{1}}=m(x-{{x}_{1}})$

$\Rightarrow y-0=3(x-0)$

$y=3x\Rightarrow 3x-y=0$

Now, equation of the normal is

$y-{{y}_{1}}=\frac{-1}{m}(x-{{x}_{1}})$

$\Rightarrow y-0=\frac{-1}{3}(x-0)$

$\Rightarrow 3y=-x\Rightarrow x+3y=0$

• Some important points to remember:
1. If$\frac{dy}{dx}>0,$ then the tangent to the curve makes an acute angle with the x-axis

1. If $\frac{dy}{dx}<0,$ then the tangent to the curve makes an obtuse angle with the x-axis.
2. If $\frac{dy}{dx}=0,$then the tangent is parallel to the x-axis.
3. If $\frac{dy}{dx}=\infty ,$i.e., $\frac{dy}{dx}=0,$ then tangent is perpendicular to the x-axis i.e.it is parallel to the y-axis.
4. If $\frac{dy}{dx}=\pm 1$ i.e. $\theta =45{}^\circ$or $135{}^\circ$, then the tangent is equally inclined to both the axis.
5. The slope of the line having the equation

$ax+by+c=0$ is written as

$m=\frac{-a}{b}=\frac{\text{coefficient of }x\text{ with negative sign}}{\text{coefficient of y}}$

1. Two lines having slopes ${{m}_{1}}$and ${{m}_{2}}$are:

(a) perpendicular if ${{m}_{1}}.{{m}_{2}}=-1$

(b) parallel if ${{m}_{1}}={{m}_{2}}$

• Angle of intersection of two curves: Let $y=f(x)$ & $y=g(x)$ are two curves intersecting at point $P(x,y).$ Then, the angle of intersection of two curves at point $P(x,y)$ is equal to the angle between the tangent of the respective curves at that point of intersection $P(x,y).$

Let ${{m}_{1}}=$ slope of the tangent to curve $y=f(x)$ at $P(x,y).$

${{m}_{2}}=$ slope of the tangent to curve $y=g(x)$ at $P(x,y)$

(In a triangle, sum of two interior angles is equal to an exterior angle.)

$\therefore \,\,\,\,\alpha +\theta =\beta$

So, $\alpha$be the angle of the tangent in the positive direction of x-axisa and $\beta$ be the angle of the tangent to curve $y=g(x)$ in the positive direction of x-axis.

$\Rightarrow \theta =\beta -\alpha \Rightarrow \tan \theta =\tan (\beta -\alpha )=\frac{\tan \beta -\tan \alpha }{1+\tan \beta .\tan \alpha }$

$\Rightarrow \tan \theta =\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}.{{m}_{1}}},\,\,\,\theta ={{\tan }^{-1}}\left| \left. \frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}.{{m}_{1}}} \right| \right.$

• Length of Tangent/Normal/Sub-tangent& subnormal:

Let the tangent and normal on the point P(x, y) at the curve meet x-axis at T and N respectively. Let M be the foot of the ordinate at P. Then

In $\Delta \text{PTM},\,\sin \theta =\frac{y}{\text{PT}}\Rightarrow PT=y\cos ec\,\theta$

1. Length of the tangent $=PT=\,\,|y.\cos ec\,\theta |\,\,=\left| \left. y.\sqrt{1+{{\cot }^{2}}\theta } \right| \right.$

$=y.\sqrt{1+\frac{1}{{{\left( \frac{dy}{dx} \right)}^{2}}}}=\frac{\sqrt[y]{1+{{\left( \frac{dy}{dx} \right)}^{2}}}}{\left( \frac{dy}{dx} \right)}$

Similarly,

Length of the normal $=PN=y.\sec \theta =y.\sqrt{1+{{\tan }^{2}}\theta }=y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}$

Length of the sub-tangent $(TM)=y.cot\theta =\frac{y}{\tan \theta }=\frac{y}{\left( \frac{dy}{dx} \right)}$

Length of the subnormal $(MN)=y.\tan \theta =\left| \left. y.\frac{dy}{dx} \right| \right.$

Solved Examples

1. A man 2 metre high walks at uniform speed of 6 metre per minute away from the lamp post of height 5 metre. Find the rate at which the length of his shadow increase.

Sol. Let AB is a lamp post. At any instant t, A man MN away x m from AB and its shadow is y metre. Given that

$AB=5m$

$MN=2m$ and $\frac{dx}{dt}=6$metre / min

$AM=x$metre and $MC=y$ metre.

Since, $\Delta ABC$ and $\Delta MNC$ be similar

So, $\frac{MN}{AB}=\frac{MC}{AC}$      $\Rightarrow \,\,\,\,\frac{AB}{MN}=\frac{AC}{MC}$

$\Rightarrow \,\,\,\,\frac{5}{2}=\frac{x+y}{y}$       $\Rightarrow \,\,\,\,\frac{5}{2}=1+\frac{x}{y}$

$\Rightarrow \,\,\,\,\frac{5}{2}-1=\frac{x}{y}$      $\Rightarrow \,\,\,\,\frac{3}{2}=\frac{x}{y}\,\,\,\,\Rightarrow 3y=2x$

Differentiating w.r.t. to t, we get

$3.\frac{dy}{dt}=2\left( \frac{dx}{dt} \right)$        $\Rightarrow \,\,\,\,\frac{dy}{dt}=\frac{2}{3}\times 6=4$m/sec

1. Use differentiation to find the approximate value of $\sqrt{0.037}.$

Sol. Let $y=f(x)=\sqrt{x}$

Let $x=0.040$ and $x+\delta x=0.037$

$\delta x=x+\delta x-x=0.037-0.040=-0.003$

For

$\because y=f(x)=\sqrt{0.04}=0.2$

$\therefore dy=\left( \frac{\delta y}{\delta x} \right).dx$

$\therefore \frac{dy}{dx}=\frac{\delta }{\delta x}(\sqrt{x})=\frac{1}{2\sqrt{x}}\le$

${{\left. \frac{dy}{dx} \right|}_{\,\,at\,\,x=0.04}}=\frac{1}{2\sqrt{x}}=\frac{1}{2\times 0.2}=\frac{1}{0.4}$

So, $\delta y={{\left( \frac{\delta y}{\delta x} \right)}_{at\,\,x=0.04}}\times \delta x=\frac{1}{0.4}\times (-0.003)=-\frac{3}{400}$

$\therefore$ Approximate value of $y=f(x)=y+\delta y=0.2+\left( \frac{-3}{400} \right)=\frac{80-3}{400}=\frac{77}{400}$

• Increasing & decreasing function (monotonic function): A function is said to be monotonic function if it is either increasing or decreasing.

• Increasing function:

(i)   A function $y=f(x)$ is said to be increasing on an interval $[a,b]$if${{x}_{1}}<{{x}_{2}}$in$(a,b)$$\Rightarrow f({{x}_{1}})\le f({{x}_{2}})$$\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)$

(ii) A function $y=f(x)$ is said to be strictly increasing on$(a,b)$if${{x}_{1}}<{{x}_{2}}$in$(a,b)$$\Rightarrow f({{x}_{1}})<f({{x}_{2}})$ for all ${{x}_{1}},{{x}_{2}}\in (a,b)$

• Decreasing function:

(i) A function $y=f(x)$ is said to be decreasing on the interval$(a,b)$ if ${{x}_{1}}>{{x}_{2}}$in$(a,b)$$\Rightarrow f({{x}_{1}})\ge f({{x}_{2}})$$\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)$

(ii) A f unction$y=f(x)$ is said to be strictly decreasing on $(a,b)$ if${{x}_{1}}>{{x}_{2}}$in$(a,b)$$\Rightarrow f({{x}_{1}})>f({{x}_{2}})$$\forall \,{{x}_{1}},{{x}_{2}}\in (a,b)$

• Test of monotonicity of function:

(i)  $f(x)$ is increasing in $[a,b]$ if $f'(x)\ge 0$$\forall x\in [a,b]$

(ii) $f(x)$ is strictly increasing in $[a,b]$if $f'(x)>0$$\forall x\in [a,b]$

(iii) $f(x)$ is decreasing in $[a,b]$ if $f'(x)\le 0$$\forall x\in [a,b]$

(iv) $f(x)$ is strictly decreasing in $[a,b]$if $f'(x)<0$$\forall x\in [a,b]$

#### Other Topics

##### Notes - Mathematics Olympiads -Application of Derivatives

You need to login to perform this action.
You will be redirected in 3 sec