# CBSE Class 11 - Mathematics - Limits and Derivatives Part-8

 Differentiation

In the previous post[Part-7], we learned about some important derivatives using the First Principle.

Let us talk about what differential co-efficient is?

Differential coefficient: Let y = f (x) be a continuous function of a variable quantity x, where x is independent and y. is a dependent variable quantity.
Let δ x be an arbitrarily small change in the value of x and δ y be the corresponding change in y then $\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}$ if it exists, is called the derivative or differential coefficient of y with respect to x and it is denoted by $\frac{dy}{dx}$, y',y1 or Dy
The process of finding the derivative of a function is called differentiation.

If we again differentiate (dy/dx) with respect to x then the new derivative so obtained is called second derivative of y with respect to x and it is denoted by $\frac{\mathrm{d}^{2}y }{\mathrm{d} x^{2}}$ or y" or y2 or D2y. Similarly, we can find successive derivatives of y which may be denoted by $\frac{\mathrm{d}^{3}y }{\mathrm{d} x^{3}}$, $\frac{\mathrm{d}^{4}y }{\mathrm{d} x^{4}}$,.............$\frac{\mathrm{d}^{n}y }{\mathrm{d} x^{n}}$,...........

DERIVATIVE OF F(X) FROM THE FIRST PRINCIPLE/AB INITIO METHOD:

If f (x) is a derivable function then, $\lim_{\delta x\rightarrow 0}\frac{\delta y}{\delta x}=\lim_{\delta x\rightarrow 0}\frac{f(x+\delta x)-{f(x)}}{\delta x}= f'(x)=\frac{dy}{dx}$ or simply $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

The derivative of sin x:

Let y = sin x, where the argument x is expressed in radians.
We have y + Δ y = sin (x + Δx). Hence, Δy=sin (x + Δx) - sin x
or, $\Delta y = 2sin\frac{\Delta x}{2}cos(x+\frac{\Delta x}{2})$
Dividing both sides of the last equality by Δx, we get
$\frac{\Delta y}{\Delta x}=\frac{2sin\frac{\Delta x}{2}}{\Delta x}cos(x+\frac{\Delta x}{2})$ or $\frac{\Delta y}{\Delta x}=\frac{2sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}cos(x+\frac{\Delta x}{2})$

Proceeding to the limit as Δx→0 and using the theorem on the limit of a product, we have
${y}'=\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\rightarrow 0}\frac{sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}.\lim_{\Delta x\rightarrow 0}cos(x+\frac{\Delta x}{2})$

From the theorem on the limit of the ratio of the sine of an infinitely small arc to the arc itself it follows that
$\lim_{\Delta x\rightarrow 0}\frac{sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}=1$

Besides, by virtue of the continuity of the function cos x, we have
$[\lim_{\Delta x\rightarrow 0}\cos(x+\frac{\Delta x}{2})]=\cos x$

Consequently, y'=1. cos x = cos x, i.e, (sin x)' = cos x

Thus, we obtain the following theorem: The derivative of sin x is equal to cos x.

Q1: Differentiate $\frac{1}{\sqrt{(x+a)}}$ from first principle.

Sol. Step 1: Let $y=\frac{1}{\sqrt{(x+a)}}=(x+a)^{-\frac{1}{2}}$

Step II: Then y + Δy = (x + Δx + a)-1/2

Step III: Then $\Delta y=[(x+\Delta x)+ a]^{-\frac{1}{2}}-(x+a)^{-\frac{1}{2}} = (x+a)^{-\frac{1}{2}}[1+\frac{\Delta x}{x+a}]^{-\frac{1}{2}} - (x+a)^{-\frac{1}{2}}$
= $(x+a)^{-\frac{1}{2}}[1-\frac{1}{2}(\frac{\Delta x}{x+a}) + \frac{(\frac{-1}{2})(\frac{-3}{2})}{2!}(\frac{\Delta x}{x+a})^{2}....]-1]$ (by the expansion of binomial theorem.)
= $(x+a)^{-\frac{1}{2}}[-\frac{\Delta x}{2(x+a)}+\frac{3(\Delta x)^{2}}{8(x+a)^{2}}.....]$

Step (IV): $\frac{\Delta y}{\Delta x}=(x+a)^{-\frac{1}{2}}[-\frac{1}{2(x+a)}+\frac{3(\Delta x)}{8(x+a)^{2}}.....]$

Step (V): Taking limit as Δx → 0
$\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}=\lim_{\Delta x\rightarrow 0}(x+a)^{-\frac{1}{2}}[-\frac{1}{2(x+a)}+\frac{3(\Delta x)}{8(x+a)^{2}}.....]$

∴ $\frac{\mathrm{d} y}{\mathrm{d} x}=(x+a)^{-\frac{1}{2}}[-\frac{1}{2(x+a)}]=-\frac{1}{2}(x+a)^{-\frac{1}{2}}=\frac{1}{2(x+a)^{\frac{3}{2}}}$

Q2: Differentiate sin-1 (ex) from the first principle.

Sol. Let y = sin-1 (ex)
Then y + Δy = sin-1 (ex + Δx )
∴ sin (y + Δy) - sin y = ex + Δx - ex
=> $2\cos \left ( y+\frac{\Delta y}{2} \right ).\sin \left ( \frac{\Delta y}{2} \right )= e^{x}\left ( e^{\Delta x }-1 \right )$
=> $2\cos \left ( y+\frac{\Delta y}{2} \right ).\frac{\sin \left ( \frac{\Delta y}{2} \right )}{\Delta y}.\frac{\Delta y}{\Delta x} = e^{x}\frac{e^{\Delta x-1}}{\Delta x}$

Now, when Δx → 0; ex + Δx → ex. Hence y + Δy →y i.e. Δy → 0
Thus applying the limit Δx→ 0 and Δy→0, we have

$\lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x\rightarrow 0} e^{x} \left ( \frac{e^{\Delta x}-1}{\Delta x} \right ).\lim_{\Delta y\rightarrow 0}\frac{1}{2\cos \left ( y+\frac{\Delta y}{2} \right )\frac{\sin \frac{\Delta y}{2}}{\Delta y}}$
i.e. $\frac{\mathrm{d} y}{\mathrm{d} x}= \lim_{x\rightarrow 0} e^{x}\frac{\left [ \Delta x +\frac{(\Delta x)^{2}}{2!} + \frac{(\Delta x)^{3}}{3!}+.....+infinity\right ]}{\Delta x} \times \lim_{y\rightarrow 0}\frac{1}{\cos \left ( y + \frac{\Delta y}{2}\right )\frac{\sin \frac{\Delta y}{2}}{\frac{\Delta y}{2}}}$
i.e. $\frac{\mathrm{d} y}{\mathrm{d} x}= e^{x}.\frac{1}{\cos y} = \frac{e^{x}}{\sqrt{1 -\sin ^{2}y}} = \frac{e^{x}}{\sqrt{1 - e^{2x}}}$

Ch 13 Limits and Derivatives (Q & A) Part-5

Ch 13 Limits and Derivatives (Q & A) Part-6

Ch 13 Limits and Derivatives (Q & A) Part-7