# Applications of Derivatives

Table of Content

There are several applications of derivatives that includes:

• Rates of Change
• Critical Points
• Minimum and Maximum Values
• Mean Value Theorem
• Linear Approximations

## Equation Of Tangents And Normal

If a tangent at any point P makes an anglewith the positive direction of the x-axis then $\frac{dy}{dx}$= tan

Equation of tangent at point P(x1y1) is

y – y1 = ${{\left( \frac{dy}{dx} \right)}_{({{x}_{1}}{{y}_{1}})}}(x-{{x}_{1}})$

Equation of normal at point P (x1y1)

Y – y1 = $-{{\left( \frac{dy}{dx} \right)}_{({{x}_{1}}{{y}_{1}})}}(x-{{x}_{1}})$

Two curves cut orthogonally:

m1 m2 = $\Rightarrow a=\pm {}^{2}/{}_{\sqrt{3}}$

### Monotonicity

Functions are said to be monotonic if they are either increasing decreasing in their entire domain eg.

f(x) = ex, f(x) = n x, f(x) = 2x + 3 are some examples

Functions which are increasing and decreasing in their domain are said to be non-monotonic

For example: f(x) = sinx , f(x) = x2

### Monotonocity Of A function At A Point

A function is said to be nonotonically decresing at x = a of f(x) satisfy

$\left. \begin{matrix} f(x+h)<f(a) \\ f(x+h)<f(a) \\ \end{matrix} \right]$ for a small positive h

f’(x) will be positive if the function is increasing

f’(x) will be negative if the function is decreasing

f’(x) will be zero when the function is at its maxima or minima

Example: Find the least value of K for which the function x2 + kx + 1 is an increasing function in the interval k x < 2

Solution: f(x) = x2 + kx + 1

For f(x) to be increasing, f1(x) > 0

$\frac{d}{dx}\left( {{x}^{2}}+kx+1 \right)>0\Rightarrow 2x+k>0\Rightarrow k>-2x$

For $x\in (1,2),$ the least value of k is

Prove that the function f(x) = log (x2+1) $-{{e}^{-x}}-1$is strictly increasing $\forall x\in R$

Solution: f(x) = $\ell n({{x}^{2}}+1)-{{e}^{-x}}+1$ $\Rightarrow \,{{f}^{1}}(x)=\frac{2x}{1+{{x}^{2}}}+{{e}^{-x}}$ $\Rightarrow \,{{f}^{1}}(x)={{e}^{-x}}+\frac{2x}{x+{}^{1}/{}_{x}}$

For x < 0, 1 <$\frac{2}{x+{}^{1}/{}_{x}}<0\,\And \,{{e}^{-x}}>1$

Hence ${{e}^{-x}}+\frac{2x}{1+{{x}^{2}}}>0$

f(x) is strictly increasing fx

### Point Of Inflection

For continuous function f(x), if f’(x0) = 0 or f’’(x0) does not exist at points where f’(x0) exists and if f”(x) changes sign when passing through x = x0 then x0 is called the point of inflection

(a) If f”(x) < 0, x (a,b) then the curve y = f(x) in concave downward

(b) if f” (x) > 0, x(a, b) then the curve y = f(x) is concave upwards in (a, b)

Example: f(x) = sinx

Solution: f’(x) = cosx

f”(x) = sinx = 0 x = nπ, n $\in$ z

Example: f(x) = 3x4 – 4x3

### Problems On Applications Of Derivatives

Example: Find the equation of the normal y = x3 – 3x which is parallel to 1x + 18y = 9

Solution: The curve is y = x3 – 3x

$\Rightarrow \frac{dy}{dx}=3{{x}^{2}}-3$

∴ $\frac{dy}{dx}=9=3{{x}^{2}}-3\Rightarrow x=\pm 2$

The required normal are x + 9y = 20 & x + 9y + 20 = 0

Example: Find the equation of all possible normal to the parabola x+ = 4g from the point (1, 2)

Solution: Let point Q be $\left( h,\frac{{{h}^{2}}}{4} \right)$ and point P be the point of contact on the curve

Now slope of normal at q, point (x1 y1)

• Find the angle between the curves y2 = 4x and y = ${{e}^{{}^{-x}/{}_{2}}}$
• Find the values of a if the curves $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{4}=1\,\,and\,\,{{y}^{3}}=16x$cut orthogonally
• Differentiating first curve $\frac{dy}{dx}=\frac{-4x}{{{a}^{2}}y}\,={{m}_{1}}$
• Differentiating second curve $\frac{dy}{dx}=\frac{16}{3{{y}^{2}}}\,={{m}_{2}}$