DERIVATIVE FORMULAS
Power Rule:
(d/dx) (xn ) = nxn-1
Derivative of a constant,
(d/dx) (a) = 0
Derivative of a constant multiplied with function f(x)
(d/dx) (a. f) = a (d/dx) f(x)
Sum Rule
(d/dx) [f (x)+ g(x)] = (d/dx)f (x)+ (d/dx) g(x)
(d/dx) [(f(x) - g(x)] = (d/dx)f(x) - (d/dx) g(x)
Product Rule
(d/dx) [f (x)*g(x)]= f(x) (d/dx)g(x) + g (x) (d/dx) f (x)
Quotient Rule
d/dx (f(x)/g(x))= g(x) d/dx f(x)-f (x)d/dx g(x)
Derivatives of some trigonometric functions
(d/dx) sin x = cos x
(d/dx) cos x = -sin x
(d/dx) tan x = sec2 x
(d/dx) cot x = - cosec2 x
(d/dx) sec x = sec x tan x
(d/dx) cosec x = - cosec x cot x
Derivatives of elementary functions
(d/dx ) log x = 1/ x
(d/dx) ex = ex
(d/dx) ax = ax. log a , where a>0, a ≠ 1
(d/dx) √x =1/(2 √x)(d/dx) log ax = 1/(log a)x, x > 0
(d/dx) k =0 , where k is a constant
Derivatives of hyperbolic trigonometric functions
(d/dx ) sin hx = cos hx
(d/dx ) cos hx = sin hx
(d/dx ) tan hx = sech2 x
(d/dx ) cot hx = -cosech2 x
(d/dx ) sec hx = -sech x tanh x
(d/dx ) cosec hx = -cosech x cot hx
Derivatives of Inverse trigonometric functions
(d/dx ) sin -1 x = 1∕√(1-x²) , -1<x<1
(d/dx ) cos -1 x = -1∕√(1-x²) , -1<x<1
(d/dx ) tanh -1 x = 1/1+x²
(d/dx ) cot -1 x = -1/1+x²
(d/dx ) cosec -1 x = - 1/|x|√x²-1 , |x|>1
Derivatives of Inverse hyperbolic functions
(d/dx ) sinh -1 x = 1/(√x²+1)
(d/dx ) cosh -1 x = 1/(√x²-1)
(d/dx ) tanh -1 x = 1/(1-x²)
(d/dx ) coth -1 x = 1/x(√1-x²)
(d/dx ) cosech -1 x = 1/x(√1+x²)
Derivative Formula
Differentiation or derivative is a process, where we find the instantaneous
rate of change in function based on one of its variables. The most
common example is the rate change of displacement with respect to time,
called velocity. The opposite of finding a derivative is anti-differentiation.
Differentiation is a method of finding the derivative of a function
If x is a variable and y is another variable, then the rate of change of x with
respect to y is given by dy/dx.
This is the general expression of derivative of a function and is represented as
f'(x) = dy/dx, where y = f(x) is any function.
In real life, the derivatives are used by the people for maintaining a reservoir ,need
to know when will a reservoir overflow knowing the depth of the water at several
instances of time, Rocket scientists need to compute the precise velocity with
which the satellite needs to shot out from the rocket knowing the height of the
rocket at various times. Financial institutions needs to predict the changes in the
value of a particular stock knowing its present value. Many such cases it is
desirable to know how a particular parameter is changing with respect to some
other parameter.
PROBLEMS USING DERIVATIVE FORMULAS
For some constants a and b, find the derivative of
(i) (x − a) (x − b)
(ii) (ax2 + b)2
Solution:
(i) (x – a) (x – b)
f(x) = (x-a)(x-b)
f(x)= x²- (a+ b)x+ ab
By differentiating on both sides
d/dx f(x) = d/dx (x²- (a+ b)x+ ab)
=d/dx x²- (a +b)d/dx(x)+ d/dx (ab)
=2x - (a+ b) *1 + 0 [ since (d/dx) (xn ) = nxn-1
and (d/dx) (a) = 0 ]
= 2x -a-b
(ii) (ax2 + b)2
f(x) = (ax2 + b)2
Differentiating on both sides by
dy/dx f(x) = d/dx (ax2 + b)2
d/dx f(x) = a d/dx ( x2 )+ d/dx( b2 )
= 4ax (ax2 + b)
Find the derivative of
(i) (5x3 + 3x – 1) (x – 1)
(ii) x-3 (5 + 3x)
Solutions
(i) (5x3 + 3x – 1) (x – 1)
f(x) = (5x3 + 3x – 1) (x – 1)
Differentiating on both sides by product rule
d/dx f (x) = d/dx (5x3 + 3x – 1) (x – 1)
= (5x3 + 3x – 1) d/dx (x-1) +(x-1) d/dx (5x3 + 3x – 1)
= (5x3 + 3x – 1) d/dx (x) - d/dx (1) +(x-1) d/dx (5x3 ) + d/dx (3x )– d/dx (1)
= (5x3 + 3x – 1) * 1- 0 +(x-1)(15x2+3)
= 5x3 + 3x – 1+15x3+3x-15x2-3
= 20 x3 -15 x2+6x-4
(ii) x-3 (5 + 3x)
solution
f(x) = x-3 (5 + 3x)
Differentiating on both sides by product rule
d/dx f(x) = d/dx [x-3 (5 + 3x)]
= x-3 d/dx(5 + 3x)+(5 + 3x) d/dx x-3
= x-3 d/dx(5) + d/dx (3x)+(5 + 3x) d/dx x-3
= x-3 (0+3)+(5+3x) (-3 x-3-1)
= 3x-3-15x-4-9x-3
= -6 x-3-15x-4
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