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The derivative of f, denoted by f'(x) is given by f'(x) = limΔx→0 (Δy)/(Δ x) = dy/dx
The right hand derivative of f at x = a is denoted by f'(a+) and is given by f'(a+) = limh→0+(f(a+h)-f(a))/h
For a function to be differentiable at x=a, we should have f'(a-) = f'(a+) i.e. limh→0 (f (a-h)-f (a))/ (-h) = limh→0 (f(a+h) – f (a))/h.
limh→0 sin 1/h fluctuates between -1 and 1.
If at a particular point say x =a, we have f'(a+) = t1 (a finite number) and f'(a-) = t2 (a finite number) and if t1 ≠ t2, then f' (a) does not exist, but f(x) is a continuous function at x = a.
Continuity and differentiability are quite interrelated. Differentiability always implies continuity but the converse is not true. This means that a differentiable function is always continuous but if a function is continuous it may or may not be differentiable.
Some basic formulae:
d/dx (logex) = 1/x, x > 0
d/dx (logax) = 1/x logea
d/dx (sin x) = cos x
d/dx (cos x) = – sin x
d/dx (tan x) = sec2x, x ≠ (2n+1) π/2, n∈ I.
d/dx (cot x) = - cosec2x, x ≠ nπ, n∈ I.
d/dx (sec x) = sec x tan x, x ≠ (2n+1) π/2, n∈ I.
d/dx (cosec x) = - cosec x cot x, x ≠ nπ, n∈ I.
d/dx (sin-1 x) = 1/√(1 – x2), -1 < x < 1
d/dx (cos-1 x) = – 1/√(1 – x2), -1 < x < 1
d/dx (tan-1 x) = 1/(1 + x2)
d/dx (cot-1 x) = - 1/(1 + x2)
d/dx (cosec-1 x) = - 1/|x|√(x2 – 1), |x| > 1
d/dx (sec-1 x) = 1/|x|√(x2 – 1), |x| > 1
d/dx (sinh x) = cosh x
d/dx (cosh x) = sinh x
d/dx (tanh x) = sech2x
d/dx (coth x) = - cosech2x
d/dx (sech x) = – sech x tanh x
d/dx (cosech x) = - cosech x coth x
If a function is not derivable at a point, it need not imply that it is discontinuous at that point. But, however, discontinuity at a point necessarily implies non-derivability.
In case, a function is not differentiable but is continuous at a particular point say x = a, then it geometrically implies a sharp corner at x = a.
A function f is said to be derivable over a closed interval [a, b] if :
For the points a and b, f'(a+) and f'(b-) exist and
For ant point c such that a < c < b, f'(c+) and f'(c-) exist and are equal.
If y = f(u) and u = g(x), then dy/dx = dy/du.du/dx = f'(g(x)) g'(x). This method is also termed as the chain rule.
For composite functions, differentiation is carried out in this way:
If y = [f(x)]n, then we put u = f(x). So that y = un. Then by chain rule:
dy/dx = dy/du.du/dx = nu(n-1)f' (x) = [f(x)](n-1) f' (x)
f(x) + g(x)
f(x) - g(x)
f(x) . g(x)
f(x) / g(x), provided g(a) ≠ 0
If the function f(x) is differentiable at x = a while g(x) is not derivable at x = a, then the product function f(x). g(x) can still be differentiable at x = a.
Even if both the functions f(x) and g(x) are not differentiable at x = a, the product function f(x).g(x) can still be differentiable at x = a.
Even if both the functions f(x) and g(x) are not derivable at x = a, the sum function f(x) + g(x) can still be differentiable at x = a.
If function f(x) is derivable at x = a, this need not imply that f'(x) is continuous at x = a.
Differentiation using substitution: The following substitutions may be used for computing the differentiation of the functions:
√a2 - x2, use x = a sin θ or a cos θ
√a2 + x2, use x = a tan θ or a cot θ
√x2 - a2, use x = a sec θ or a cosec θ
√(a + x) or √(a – x) , use x = a cos 2θ
√2ax - x2, use x = a (1 – cos θ)
If x = f(t) and y = g(t), where t is a parameter, then dy/dx = (dy/dt)/(dx/dt) = g’(t)/f’(t)
If a function is in the form of exponent of a function over another function such as f(x)g(x), we first take logarithm and then differentiate.
Let y = f(x) and z = g(x), then the differentiation of y with respect to z is dy/dz = dy/dx / dz/dx = f’(x)/g’(x)
Leibnitz Theorem:
If u and v are two functions of x possessing nth order derivatives, then
Dn(u.v) = nC0(Dnu)v + nC1(Dn-1u)(Dv) + nC2(Dn-2u)(D2v) + ….….. ….…. + nCr(Dn-ru)(Drv) + ….…. + nCn(Dnv)
Partial coefficient of f(x,y) with respect to x is the ordinary differential coeffiicient of f(x, y) when y is regarded as a constant.
If f”(x, y) be a function of two variables such that ∂f/∂x and ∂f/∂y both exist then we can talk of second order partial derivatives as well.
The partial derivative of ∂f/∂x with respect to x is fxx i.e. ∂2f/∂x2
The partial derivative of ∂f/∂x with respect to y is fxy i.e. ∂2f/∂x∂y
The partial derivative of ∂f/∂y with respect to x is fyx i.e. ∂2f/∂y∂x
The partial derivative of ∂f/∂y with respect to y is fyy i.e. ∂2f/∂y2
Note: ∂2f/∂x∂y = ∂2f/∂y∂x.
Let f(x, y) be a homogeneous function of order n so that f(tx, ty) = tnf(x, y), then
xi ∂f/∂xi = n f(x)
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