derivative of function| calculus,instantaneous rate of change

Derivative of function

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Derivative of function : In the earlier section we have learnt the slope of a tangent line using the limit. Here you will learn the derivative of a function.
The limit used to define the slope of a tangent line is also used to define one of the two fundamental operations of calculus—differentiation.

Definition of the derivative of the function
The derivative of at is

f'(x) = $\frac{\text{d}y}{\text{d}x} =\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

provided the limit exists. For all for which this limit exists, f' is a function of x

This definition of derivative of a function is used to find the instantaneous rate of change at a particular value of x. The process of finding the derivative of a function is called the
Differentiation.

A function is differentiable at when its derivative exists at and is differentiable on an open interval (a,b) when it is differentiable at every point in the interval.

The other notation for f'(x) is 1) y' 2) $\frac{\text{d}y}{\text{d}x}$ 3) $\frac{\text{d}[f(x)]}{\text{d}x}$ 4) $D_{x}[y]$

Examples on finding the derivative of function using the limit process

Example 1: To find the derivative of function f(x) = $x^{3}$ + 2x using the definition of derivative or using the limit process.
Solution :

f'(x) = $\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

Now in the given function replace x by (x + $\triangle x$) and simplify

=$\lim_{\triangle x \rightarrow 0}\frac{[(x+\triangle x)^{3}+2(x + \triangle x)-(x^{3} + 2x)}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{(x^{3} + 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3}+2x + 2\triangle x)-(x^{3} + 2x)}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{x^{3} + 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3}+2x + 2\triangle x- x^{3} - 2x}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{ 3 x^{2}\triangle x + 3 x (\triangle x)^{2} + (\triangle x)^{3} + 2\triangle x}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0}\frac{ \triangle x [ 3 x^{2} + 3 x (\triangle x) + (\triangle x)^{2} + 2}{\triangle x}$

=$\lim_{\triangle x \rightarrow 0} {3 x^{2} + 3 x (\triangle x) + (\triangle x)^{2} + 2}$

Plug in $ \triangle x$= 0

= $3 x^{2} + 3 x (0) + (0)^{2} + 2$

=$3 x^{2} +2 $

Example 2: To find the derivative of f(x) = $\sqrt{x}$ using the definition of derivative or using the limit process.
Solution :

f'(x) = $\lim_{\triangle x \rightarrow 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$

Now in the given function replace x by (x + $\triangle x$) and simplify

=$\lim_{\triangle x \rightarrow 0}\frac{[(\sqrt{x + \triangle x})-(\sqrt{x})}{\triangle x}$

Now we will rationalize the numerator

= $\lim_{\triangle x \rightarrow 0}[\frac{(\sqrt{x + \triangle x})-(\sqrt{x})}{\triangle x}][\frac{(\sqrt{x + \triangle x})+(\sqrt{x})}{(\sqrt{x + \triangle x})+(\sqrt{x})}]$

= $\lim_{\triangle x \rightarrow 0}\frac{({x + \triangle x})-(x)}{\triangle x (\sqrt{(x + \triangle x)+ x})}$

= $\lim_{\triangle x \rightarrow 0}\frac{ \triangle x}{\triangle x (\sqrt{(x + \triangle x)+ x})}$

=$\lim_{\triangle x \rightarrow 0}\frac{ 1}{(\sqrt{(x + \triangle x)+ x})}$

Plug in $\triangle x $= 0

=$\frac{1}{\sqrt(x + 0)+\sqrt {x}}$

= $\frac{ 1}{\sqrt{x} + \sqrt{x}}$

=$\frac{ 1}{2\sqrt{x}}$

12th grade math

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