Differentiation, Chapter Notes, Class 12, Maths (IIT) - JEE PDF Download

Differentiation

A. Derivative Using First Principle /ab Initio Method

If f(x) is a derivable function then,

Ex.1 Using first principles, find the derivative of the function y =-cot x- x.

Sol.

Ex.2 Find by first principle the derivative of   w.r.t. x.

Sol.

Ex.3 If f (x) = (ln x)^x , find f ' (x) from the first principle.

Sol.

Derivative Of Standard Functions :

B. Rules of Differentiation

i.e. multiply the differential coefficient of each separate function by the product of all the remaining functions and add up all the results; the sum will be the differential coefficient of the product of all the functions.

Ex.4 Differentiate y = sin(x²)

Sol. If y = sin(x²), then the outer function is the sine function and the inner function is the squaring function, so the Chain Rule gives

Ex.5 If F(x) = (x² + 2)³, compute F'(x). One way to do this problem is to expand (x² + 2)³ and use the differentiation formulae.

Sol. F(x) = (x² + 2)³ = x⁶ + 6x⁴ + 12x² + 8,

F'(x) = 6x⁵ + 24x³ + 24x.

Another method uses the Chain Rule. Let g and f be the functions defined, respectively, by g(x) = x² + 2 and f(y) = y³. Then f(g(x)) = (x² + 2)³ = F(x), and, according to the Chain Rule,

F'(x) = [f(g(x))]' = f'(g(x)) g'(x).

Since g'(x) = 2x and f'(y) = 3y², we get f'(g(x)) = 3(x² + 2)² and

F'(x) = 3(x² + 2)² (2x) = 6x (x⁴ + 4x² + 4),

which agrees with the alternative solution above.

Let f and g be two differentiable functions. The formation of the composite function f(g) is suggested by writing u = g(x) and y = f(u). Thus x is transformed by g into u, and the resulting u is then transformed by f into y = f(u) = f(g(x)). We have

The idea that one can simply cancel out du in (1) is very appealing and accounts for the popularity of the notation. It is important to realize that the cancellation is valid because the Chain Rule is true, and not vice versa. Thus, du is simply a part of the notation for the derivative and means nothing by itself. Note also that (1) is incomplete in the sense that it does not say explicitly at what points to evaluate the derivatives. We can add this information by writing

Ex.6 

Sol. 

Ex.7 Differentiate g(x) =

Sol.  is the outer function.

Ex.8 If y = , find .

Sol.

Ex.9  

Sol.

C. Logarithmic Differentiation

To find the derivative of :

(i) a function which is the product or quotient of a number of functions OR

(ii) a function of the form [f(x)]^g(x) where f & g are both derivable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic Differentiation.

Steps in Logarithmic Differentiation

1. Take natural logarithms of both sides of an equation y = f(x) and use the Laws of Logarithms to simplify.

2. Differentiate both sides with respect to x.

3. Solve the resulting equation for y'.

If f(x) < 0 for some values of x, then in f(x) is not defined, but we can write |y| = |f(x)| and

Ex.10 Find

Sol.

Ex.11 Differentiate y = .

Sol. Using logarithmic differentiation, we have

Ex.12 Find , where y = (x + 1)^2x.

Sol.

is the sum of the two special forms and therefore we may, instead of taking logarithms in any particular example, consider first u constant and then v constant and add the results obtained on these suppositions.

D. Parametric Differentiation

E. Differentiation of Implicit Functions

Assume that the equation F(x, y) = 0 specifies y as an implicit function of x. In what follows we shall consider this function to be differentiable.

Differentiating both sides of the equation F(x, y) = 0 with respect to x, we obtain a first-degree equation with respect to y'. This equation easily yields y', that is, the derivative of the implicit function.

Ex.13 Find from the equation x³ + ln y- x²e^y = 0.

Sol. Differentiating both sides of the equation with respect to x, we obtain

Steps for Implicit Differentiation

1. Differentiate both sides of the equation with respect to x.

2. Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation.

3. Factor dy/dx out of the left side of the equation.

4. Solve for dy/dx by dividing both sides of the equation by the left-hand factor that does not contain dy/dx.

Ex.14 Find dy/dx given that y³ + y²- 5y- x² =-4.

Sol.

Ex.15 Find if xy- x = 1.

Sol. We must find at x = 1. Assume y is a function of x,y = y(x). The relation now is xy (x)- x = 1.

Remark If, in the preceding example, we had explicitly solved for y, we would have obtained

To see that this is exactly the result obtained by implicit methods, note that if 1 + 1/x is substituted for y ⇒

Ex.16 Find dy/dx if 2x² + xy- 3y² = x.

Sol. If we assume y is a function of x and represent it by y(x), the equation reads 2x² + xy(x)- 3[y(x)]² = x.

Be careful : When applying Formula (2), keep in mind that the only values of x and y that can be substituted into the right-hand side of Formula (2) are those values that satisfy the original condition 2x² + xy- 3y² = x. for instance, we might substitute x = 1, y = 2 to obtain (dy/dx) = (1- 4- 2)/(1- 12) = 5/11; however, (1, 2) is not a point on the graph 2x² + xy- 3y² = x, so the calculation of dy/dx at this point is totally meaningless.

The equation in Example can be written as -3y² + xy + (2x²- x) = 0 and hence is a quadratic equation in y (an equation of the form Ay² + By + C = 0 where A =-3, B = x and C = 2x²- x). Hence, we could use the quadratic formula to solve this equation for y in terms of x, obtaining

Though we are able to find y explicitly in terms of x, the resulting expression is fairly complex, and it still might be best to find dy/dx implicitly as in Example.

Warning : It is important to realize that implicit differentiation is a technique for finding dy/dx that is valid only if y is a differentiable function of x, and careless application of the technique can lead to errors. For example, there is clearly no real-valued function y = f(x) that satisfies the equation
x² + y² =-1, yet formal application of implicit differentiation yields the :derivative: dy/dx =-x/y. To be able to evaluate this "derivative," we must find some values for which x² + y² =-1. Because no such values exist, the derivative does not exist.

Ex.17 If x³ + y² = sin (x + y), find

Sol. given, x³ + y³ = sin (x + y) ...(i)

differentiating w. r. t. x, we get

Ex.18 If x = , prove that = 2x² + y²- 3xy

Sol.

x =

F. Derivative of Inverse Functions

If the inverse functions f & g are defined by y = f(x) & x = g(y) & if

The truth of this is also manifested geometrically, for and are respectively the tangent and the co-tangent of the angle y which the tangent to the curve y = f(x) makes with the x-axis.

This formula is very useful in the differentiation of an inverse function.

Ex.19 Find the derivative of   and mention the points of non differentiability.

Sol.

f is continuous for all x but not diff. at x = 1 ,-1

G. Derivative of a Determinant

Ex.20 Let f(x) = . Show that f '' (x) = 0 and that f(x) = f(0) + k x where k denotes the sum of all the co-factors of the elements in f(0).

Sol. 

H. Higher order Derivatives

Let a function y = f(x) be defined on an open interval (a, b) . It's derivative, if it exists on (a, b) is a certain function f '(x) [or (dy/dx) or y ] & is called the first derivative of y w. r. t. x.

If it happens that the first derivative has a derivative on (a , b) then this derivative is called the second derivative of y w. r. t. x & is denoted by f ''(x) or (d²y/dx²) or y '' .

Once we have found the derivative f' of any function f, we can go on and find the derivative of f'.

Similarly, the 3^rd order derivative of y w. r. t. x , if it exists, is defined by .

It is also denoted by f '''(x) or y '''.

The third derivative, written f''', is the derivative of the second derivative, and, in principle, we can go on forever and form derivatives of higher order. We adopt the alternative notation f⁽ⁿ⁾ for the nth derivative of f.

Ex.21 Find y'' if x⁴ + y⁴ = 16.

Sol. Differentiating the equation implicitly with respect to x, we get 4x³ + 4y³y' = 0

Solving for y' gives y' = -

to find y'' we differentiate this expression for y' using the Quotient Rule and remembering that y is a function of

If we now substitute Equation 1 into this expression, we get

But the values of x and y must satisfy the original equation x⁴ + y⁴ = 16. So the answer simplifies to

Ex.22 If y = x sin x, prove that  = 0

Sol.

Ex.23 A function f(x) is so defined that for al x, [f(x)]ⁿ = f(nx). Prove that f(x) . f'(nx) = f'(x) . f(nx), where f'(x) denotes derivative of f(x) w.r. to x.

Sol. Given [f(x)]ⁿ = f(nx)                                                                 .....(1)

differentiating both sides w. r to x, we get

n[f(x)]^{n- 1} . f'(x) = f'(nx) . (n . 1)             or,                   [f(x)ⁿ . f'(x) = f'(nx)

multiplying both sides by f'(x), we get

[f(x)ⁿ . f'(x) = f'(nx) . f(x) or, f(nx) . f'(x) = f'(nx) . f(x)                           [from (1)]

Ex.24 y =  , then prove that, = x² + 4.

Sol.

I. L' Hopital's Rule

Let f and g be functions that are differentiable on an open interval (a, b) containing c, except possibly at c itself. Assume that g'(x) 0 for all x in (a, b), except possibly at c itself. If the limit of f(x)/g(x) as x approaches c produces the indeterminate frorm 0/0, then

provided the limit on the right exists (or is infinite). This result also applies if the limit of f(x)/g(x) as x approaches c produces any one of the indeterminate forms

∞ / ∞,     (- ∞) / ∞   , ∞ / (- ∞).

Ex.25                      L' Hopital's Rule.

Sol. .

Ex.26 Determine

Sol.

Ex.27 Find

Sol. The inconvenience of continuously differentiating the denominator, which involves tan² x as a factor, may be partially avoided as follows. We write

To evaluate the limit on the R.H.S., we notice that the numerator and denominator both become 0 for x = 0.

Ex.28 Find the constants 'a' (a > 0) and 'b' such that ,  = 1.

Sol.

Ex.29 Evaluate

Sol.

Ex.30 Evaluate :

Sol.