Derivatives of some trignometric function and rules of differentiatiom

inlogicalmath May 26, 2021

                           Differentiation 

Introduction :-

The history of mathematics presents the development of calculus as being accredited to sir issac Newton 
(1642-1727) an English physicst and mathematics and Gotifried Wilhelm Leibnitz(1646-1716) A German physicst and mathematician. The Derivative is one of the fundamental ideas of calculus. 

It all about rate of change in a function. We try to find interpretations of these changes in a mathematical way. The symbol `\delta` will be used to respresent the change, for example `\delta x` represents a small change in the variable x and it is read as "change in x" or "increment in x ". 

`\delta y` is the corresponding change in y is a function of x.

We have already studied studied the basic concept, derivatives of derivatives of standard functions and rules differentiation in previous standard. 

This year, in this chapter we are going to study the geometrical meaning of derivtive, derivatives of composite, Inverse of Composite, Inverse, Logarithmic, Implicit and parametric function and also higher order derivatives.We also add some more rules of differentiation.

Differentiation

 Derivatives of f(x) with respective to x, at x=a is given by

$$ f^{'}(x)=\lim_{x \to 0} \frac{f(a+h)-f(a)}{h}$$

  •  The derivative can also be defined for f(x) at any point x on the open interval as $$ f^{'}(x)=\lim_{x \to 0} \frac{f(a+h)-f(a)}{h}$$

If the function is given as y=f(x) then its derivative is written as

$\frac{dy}{dx}=f^{'}(x)$.

  • For a differentiable function y=f(x) if $\delta x$ is a small increment in x and the corresponding increment in y is $\delta y$ then $$\lim_{\delta x \to 0}\frac{\delta y}{\delta y}=\frac{dy}{dx}.$$
  •  Derivatives of some standard functions.

Rules of differentiation

if u and v are differentiable function of x such that
(i)y=u+v then $\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
(ii)y=u-v then $\frac{dy}{dx}=\frac{du}{dx}-\frac{dv}{dx}$
(iii)y=uv= then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$
(iv)y=$\frac{u}{v}$ where $v\neq0$ then
 $\frac{dy}{dx}$=$\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$


    
rules of differentiation



derivatives of function







Tags: