Calculus: Differentiation Flashcards
What is the differential of y = axn?
if y = axn, then dy/dx = anxn-1
also
if f(x) = axn, then f ‘(x) = anxn-1
Differentiate y = x3.sinx with respect to x.
If y = x3·sinx
dy/dx = x3(cos x) + sinx(3x2)
= x3cosx + 3x2sinx
= x2(xcosx + 3sinx)
What is the differential of y = ax?
If y = ax, then dy/dx = a
also
if f(x) = ax, then f ‘(x) = a
What is the differential of y = a?
If y = a, then dy/dx = 0
also
if f(x) = a, then f ‘(x) = 0
What is the second derivative of y = axn
If y = axn, then dy/dx = anxn-1
and
d2y/dx2 = (n-1)anxn-2
e.g.
if y = 2x3, then dy/dx = 6x2
and
d2y/dx2 = 12x
Note: y = f(x)
dy/dx = f ‘(x)
d2y/dx2 = f ‘‘(x)
Given the equation y = f(x) and a point, P, on the curve with co-ordinates (x1, y1), find the equation of the tangent to the curve.
The equation of a line can be shown to be:
(y - y1) = m(x - x1)
where m = gradient of the line.
The gradient is found using differentiation.
m = dy/dx
Given the equation y = f(x) and a point, P, on the curve with co-ordinates (x1, y1), find the equation of the normal to the curve.
The equation of a line normal to the tangent can be shown to be:
(y - y1) = (-1/m)(x - x1)
where m = gradient of the line tangent to the curve.
This gradient is found using differentiation.
m = dy/dx
What is the derivative of Sin x?
If y = Sin x, then dy/dx = cos x
What is the derivative of Cos x?
if y = Cos x, then dy/dx = - Sin x
What is the derivative of ex with respect to x?
If y = ex, then dy/dx = ex
Differentiate y = uv , where u and v are functions of x.
Hint: Product rule.
If y = uv , then dy/dx = u(dv/dx) + v(du/dx)
i.e. To differentiate a product of two functions:
Put down the first (differentiate the second) + put down the second (differentate the first)
Differentiate y = x4.cos x with respect to x.
If y = x4 cos x , then dy/dx = x4(-sin x) + cosx(4x3)
= 4x3cosx - x4sinx
=x3(4cosx - xsinx)
Differentiate y = x5.ex with respect to x.
If y = x5 ex, then dy/dx = x5(ex) + ex(5x4)
= x5ex +5x4ex
=x4ex(x +5)
Differentiate y = ex.sinx with respect to x
If y = ex.sinx, then dy/dx = ex(cosx) + sinx(ex)
= excosx + exsinx
=ex(cosx + sinx)
Differentiate y = 4x3.sinx with respect to x
If y = 4x3.sinx, then dy/dx = 4x3(cosx) + sinx(12x2)
= 4x3cosx + 12x2sinx
= 4x2(xcosx + 3sinx)
Differentiate y = ex.cosx with respect to x
If y = ex.cosx, then dy/dx = ex(-sinx) + cosx(ex)
= excosx - exsinx
= ex(cosx - sinx)
Differentiate y = cosx.sinx with respect to x
If y = cosx.sinx, then dy/dx = cosx(cosx) + sinx(-sinx)
= cos2x - sin2x
= cos(2x)
Differentiate y = 3x3.ex with respect to x
If y = 3x3.ex, then dy/dx = 3x3(ex) + ex(9x2)
= 3x3ex + 9x2ex
= 3x2ex(x + 3)
Differentiate y = 2x5.cosx with respect to x
If y = 2x5.cosx, then dy/dx = 2x5(-sinx) + cosx(10x4)
= 10x4cosx - 2x5sinx
=2x4(5cosx - xsinx)
Differentiate y = u/v , where u and v are functions of x.
Hint: Quotient rule.
To differentiate a quotient of two functions:
[Put down the bottom (differentiate the top) - put down the top (differentiate the bottom)] all over the bottom squared.
“low d high over high d low, all over the square of what’s below”

What is the derivative of tan x?
Hint: Quotient Rule
Tan x = sin x / cos x
If y = sinx / cosx,
then dy/dx = [cos x.cos x - sin x.(-sin x)] / (cos x)2
= (cos2x + sin2x) / cos2x
= 1 / cos2x
= sec2x
Differentiate y = sin x / x2 with respect to x.
If y = sin x / x2, then dy/dx = [x2(cosx) - sinx (2x)] / (x2)2
= [x(xcos x - 2sin x)] / x4
= (xcos x - 2sin x) / x3
Differentiate y = 5ex / cos x with respect to x.
If y = 5ex / cos x,
then dy/dx = [cos x.(5ex) - 5ex.(-sinx)] / (cos x)2
= [5ex.(cos x + sin x)] / cos2 x
Differentiate y = cos x / sin x with respect to x.
If y = cos x / sin x,
then dy/dx = [sin x.(-sin x) - cos x.(cos x)] / (sin x)2
= (-sin2 x - cos2 x) / sin2 x
= - (sin2 x + cos2 x) / sin2 x
= - 1 / sin2 x
= - cosec2 x
Differentiate y = 3x2/cos x with respect to x
If y = 3x2/cos x,
then dy/dx = [cos x.(6x) - 3x2.(-sin x)] / (cos x)2
= [3x (2 cos x + x sin x)] / cos2 x
Differentiate y = tan x / ex with respect to x
If y = tan x / ex
then dy/dx = [ex (sec2 x) - tan x (ex)] / (ex)2
= [ex (sec2 x - tan x)] / (ex)2
= (sec2 x - tan x) / ex
Differentiate f (x) = g (h (x)) with respect to x
If f (x) = g (h (x)),
Then f ‘(x) = (g (h (x)))’
= g ‘(h (x)) h’ (x)
or
dy/dx = dy/du x du/dx