Mathematics Exponential Functions,Logarithmic Functions And Logarithmic Differentiation For CBSE-NCERT
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class 12 chapter 5 - continuity & Differentiability

Exponential Functions,Logarithmic Functions And Logarithmic Differentiation For CBSE-NCERT

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Topic covered

`star` Exponential Functions
`star` Logarithmic Functions
`star` Logarithmic differentiation

Exponential Functions

● Definition : The exponential function is with positive base `b > 1` is the function `y = f(x) = b^x`

● The graph of `y = 10^x` is given in the Fig

(1) Domain of the exponential function is R, the set of all real numbers.
(2) Range of the exponential function is the set of all positive real numbers.
(3) The point `(0, 1)` is always on the graph of the exponential function (this is a restatement of the fact that `b^0= 1` for any real b > 1).
(4) Exponential function is ever increasing; i.e., as we move from left to right, the graph rises above.
(5) For very large negative values of x, the exponential function is very close to 0. In other words, in the second quadrant, the graph approaches x-axis (but never meets it).

`=>color{orange}{"Exponential function with base 10 is called the common exponential function."}`
`color{orange}{"And series with base " e "instead of " 10 "This is called natural exponential function."}`


Theorem :


`1 . color{blue}{"The derivative of " \ \ a^x \ \ "w.r.t.," \ \x \ \ is \ \a^x.log_ea}` ;

`=>d/(dx) (a^x) =a^x log_ea`

`1 .color{blue}{"The derivative of " \ \ e^x \ \ "w.r.t.," \ \x \ \ is \ \e^x}` ;

`=>d/(dx) (e^x) =e^x`
Q 3115723660

Is it true that `x = e^(log x)` for all real x?
Class 12 Chapter 5 Example 28
Solution:

First, observe that the domain of log function is set of all positive real numbers.
So the above equation is not true for non-positive real numbers. Now, let `y = e^(log x)`. If
y > 0, we may take logarithm which gives us log `y = log (e^(log x) ) = log x * log e = log x`. Thus
y = x. Hence `x = e^(log x)` is true only for positive values of x.


One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation. This is captured
in the following theorem whose proof we skip.

Logarithmic Functions

`color{red}{x → log_b x = y}` if `b^y = x`

● This function, called the logarithmic function, is defined in `log_b : R^+ → R`

● As before if the base `b = 10,` we say it is common logarithms and if `b = e,` then we say it is natural logarithms.
Often natural logarithm is denoted by `ln.`



Graph of `log`

Properties :

1. `a^(log_(a) x) = x ; a ne 0 , ne 1 , x > 0`

2. `a^(log_(b) x ) = x^(log_b a) ; a , b > 0 , ne 1 , x > 0`

3. `log_a a =1, a > 0 , ne 1`

4. `log_a x = 1/(log_x a) ; x , a > 0 , ne 1`

5. `log_a x = (log_b x)/( log_b a) ; a , b > 0 , ne 1 , x > 0`

6. For `m, n > 0` and `a > 0,` `ne 1`, then

(i) `log_a (m * n) = log_a m + log_a n`

(ii) `log_a (m/n) = log_a m - log_a n`

(iii) `log_a (m^n) = n log_a m`


`color{blue}{"Theorem"}` : `color{green}{"The derivative of log" \ \ x \ \ w.r.t., x\ \ i s\ \ 1/x}`

`=>` `d/(dx) (log x) =1/x`
Q 3135723662

Differentiate the following w.r.t. x:

(i) `e^(-x)`

(ii) `sin (log x) , x > 0`

(iii) `cos^(-1) (e^x)`

(iv) `e^(cos x)`

Class 12 Chapter 5 Example 29
Solution:

(i) Let `y = e^(– x)`. Using chain rule, we have

`(dy)/(dx) = e^(-x) * d/(dx) (-x) = - e^(-x)`

(ii) Let `y = sin (log x)` . Using chain rule, we have

`(dy)/(dx) = cos (log x) *d/(dx ) (log x) = (cos (log x) )/x`

(iii) Let `y = cos^(–1) (e^x)`. Using chain rule, we have

`(dy)/(dx) = (-1)/( sqrt ( 1- (e^x)^2 ) ) * d/(dx) (e^x) = (-e^x)/(sqrt (1- e^(2x) ) )`

(iv) Let `y = e^(cos x)` . Using chain rule, we have

` (dy)/(dx) = e^(cos x) * (- sin x) = - ( sin x ) e^(cos x)`

Logarithmic Differentiation :

● In this section, we will learn to differentiate certain special class of functions given in the form

`=>y= f(x) =[ u(x)]^(v(x))`

`=>` By taking logarithm (to base e) the above may be rewritten as

`=> log y = v(x) log [u(x)]`

`=>` Using chain rule we may differentiate this to get

`1/y * (dy)/(dx) = v(x) * 1/(u(x)) * u'(x) + v'(x) * log [u(x)]`

which implies that

`color{orange}{(dy)/(dx) =y [(v(x))/(u(x)) * u'(x) + v'(x) * log[ u(x)]]}`

Note :`f (x)` and `u(x)` must always be positive as otherwise their logarithms are not defined.
Q 3145723663

Differentiate ` sqrt ( ( (x-3) (x^2 +4) )/ ( 3x^2 + 4x +5) )` w.r.t. x.
Class 12 Chapter 5 Example 30
Solution:

Let `y = sqrt ( ( (x-3) (x^2 +4) )/( 3x^2 +4x +5) )`

Taking logarithm on both sides, we have

`log y = 1/2 [ log (x-3) + log (x^2 +4) -log (3x^2 +4x +5) ]`

Now, differentiating both sides w.r.t. x, we get

`1/y * (dy)/(dx) = 1/2 [ 1/(x-3) + (2x)/(x^2 +4) - (6x+4)/( 3x^2 +4x +5) ]`

or `(dy)/(dx) = y/2 [ 1/(x-3) + (2x)/(x^2 +4) - (6x+4)/( 3x^2 +4x +5) ]`

`=1/2 sqrt (( (x-3) (x^2 +4) )/( 3x^2 +4x +5) ) [1/(x-3) + (2x)/(x^2 +4) - (6x+4)/(3x^2 +4x + 5 ) ]`
Q 3165723665

Differentiate `a^x` w.r.t. x, where a is a positive constant.
Class 12 Chapter 5 Example 31
Solution:

Let `y = a^x`. Then

`log y = x log a`

Differentiating both sides w.r.t. x, we have

`1/y (dy)/(dx) = log a`

or ` (dy)/(dx ) = y log a`

Thus ` d/(dx) (a^x) = a^x log a`
Q 3175723666

Differentiate `x^(sin x) , x > 0` w.r.t. x.
Class 12 Chapter 5 Example 32
Solution:

Let `y = x^(sin x)` . Taking logarithm on both sides, we have

`log y = sin x log x `

Therefore ` 1/y * (dy)/(dx) = sin x d/(dx) (log x) + log x d/(dx) (sin x)`

or `1/y (dy)/(dx) = (sin x) 1/x + log x cos x`

or ` (dy)/(dx ) = y [ (sin x)/x + cos x log x ] `

`= x^(sin x) [ (sin x)/x + cos x log x]`

`= x^(sin x-1) * sin x + x^(sin x) * cos x log x`
Q 3185723667

Find ` (dy)/(dx) ` , if `y^x + x^y + x^x = a^b`.
Class 12 Chapter 5 Example 33
Solution:

Given that `y^x + x^y + x^x = a^b`.

Putting `u = y^x, v = x^y` and `w = x^x`, we get `u + v + w = a^b`

Therefore ` (du)/(dx) + (dv)/(dx) + (dw)/(dx) = 0` ........(1)

Now, `u = y^x`. Taking logarithm on both sides, we have

`log u = x log y`

Differentiating both sides w.r.t. x, we have

`1/u * (du)/(dx) = x d/(dx) (log y ) + log y d/(dx) (x)`

`= x 1/y * (dy)/(dx) + log y * 1`

so `(du)/(dx) = u (x/y (dy)/(dx) + log y) = y^x [x/y (dy)/(dx) + log y ]` .... (2)


Also `v = x^y`

Taking logarithm on both sides, we have

`log v = y log x`

Differentiating both sides w.r.t. x, we have

`1/v * (dv)/(dx) = y d/(dx) (log x) + log x (dy)/(dx)`

` = y * 1/x + log x * (dy)/(dx)`

So ` (dv)/(dx) = v [ y/x + log x (dy)/(dx) ]`

` = x^y [y/x + log x (dy)/(dx) ]` ................(3)

Again `w = x^x`

Taking logarithm on both sides, we have

`log w = x log x`.

Differentiating both sides w.r.t. x, we have


`1/w* (dw)/(dx) = x d/(dx) (log x) + log x * d/(dx) (x)`

`= x * 1/x + log x * 1`

i.e. ` (dw)/( dx) = w (1+ log x)`

`= x^x (1+ log x)` ...................(4)


From (1), (2), (3), (4), we have

`y^x ( x/y (dy)/(dx) + log y) + x^y (y/x + log x (dy)/(dx) ) + x^x (1+ log x) = 0`

or ` ( x * y^(x-1) + x^y * log x) (dy)/(dx) = - x^x (1+ log x ) - y * x^(y-1) - y^x log y`

Therefore ` (dy)/(dx) = ( - [ y^x log y + y * x^(y-1) + x^x (1+ log x) ] )/( x * y^(x-1) + x^y log x)`
 
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