Friday 25 March 2022

Methods of Differentiation - CBSE Class 11 - Mathematics - Limits and Derivatives Part-11 #class11Maths #limits #calculus #differentiation #eduvictors

Methods of Differentiation 

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

Methods of Differentiation - CBSE Class 11 - Mathematics - Limits and Derivatives Part-11 #class11Maths #limits #calculus #differentiation #eduvictors

#class11Maths  #calculus 

Working rule for finding the derivative of implicit functions

First method:

Differentiate every term of f (x, y) = 0 with respect to x.

Collect the coefficients of $ \frac{\mathrm{d} y}{\mathrm{d} x} $ and obtain the value of $ \frac{\mathrm{d} y}{\mathrm{d} x} $

Second method

f (x, y) = constant, then $ \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{{\partial f}/{\partial x}}{{\partial f}/{\partial y}} $ are partial differential coefficients of f (x, y) with respect to x and y respectively.

Note: Partial differential coefficient f (x, y) with respect to x means the ordinary differential coefficient of f (x, y) with respect to x keeping y constant.


Let              y = f (x),              .........(1)

be a differentiable function with respect to the argument x in some interval (a, b). If in equation (1) we regard y as the argument and x as the function, then this new function

x = ⲫ (y)

where f[ⲫ(y)] ≡ y, is called inverse with respect to the given function.

Our task is to know the derivative y'ₓ = $\lim_{\Delta x\rightarrow 0} \frac{\Delta y}{\Delta x}$ of the inverse function x = ⲫ (y), assuming that the latter exists and is continuous on the corresponding interval without resolving equation (1).

Theorem: For a differentiable function whose derivative is not equal to zero the derivative of the inverse function is equal to the inverse of the derivative of the given function.


If the relation between the variables x and y is given by an equation containing both, and this equation is not immediately solvable for y, then y is called an implicit function of x.

Implicit functions are given by  ⲫ(x,y) = 0.

(i) In order to find dy/dx, in the case of implicit functions, we differentiate each term with respect to x regarding y a function of x and then collect terms in dy/dx together on one side to finally find dy/dx.

(ii) In answers of dy/dx in the case of implicit functions, both x and y are present.


If differentiation of an expression or an equation is done after taking log on both sides, then it is called logarithmic differentiation. This method is useful for the function having following forms:

When base and power both are the functions of x i.e. the functions is of the form $[f(x)]^{g(x))}$.

$y = [f(x)]^{g(x))}$

$\log y = g(x)\log[f(x)]$

$\frac{1}{y}.\frac{dy}{dx}= \frac{d}{dx}[g(x)\log[f(x)]]$

$\frac{dy}{dx}= [f(x)]^{g(x))}.\left \{ \frac{d}{dx}[g(x)\log[f(x)]] \right \}$


Let $y = a^x$, where a > 0

Then ln y = x ln a

Taking the derivative of both sides with respect to x, we have

$\frac{1}{y}y' = \ln a$

Hence, y' = y ln a,

or finally, $(a^x)' = a^x \ln a $ ......... (1)

Thus the derivative of an exponential function is equal to the function itself multiplied by naturl logarithm of the base.

Using chain rule, we can write the more general form of this as follows:

The derivative of $a^{f(x)}:\frac{d}{dx}(a^{f(x)}) = a^{f(x)}f'(x) \ln a $

In the next post, we'll talk about differentiation by trigonometric substitutions.


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