# First Derivative Formulas

y is a function y = y(x)
C = constant, the derivative(y') of a constant is 0

y = C => y' = 0

Example: y = 5, y' = 0

If y is a function of type y = xn the derivative formula is:

y = xn => y' = nxn-1

Example: y = x3 y' = 3x3-1 = 3x2
y = x-3 y' = -3x-4

From the upper formula we can say for derivative y' of a function y = x = x1 that:

if y = x then y'=1
y = f1(x) + f2(x) + f3(x) ...=>
y' = f'1(x) + f'2(x) + f'3(x) ...

This formula represents the derivative of a function that is sum of functions.
Example: If we have two functions f(x) = x2 + x + 1 and g(x) = x5 + 7 and y = f(x) + g(x) then y' = f'(x) + g'(x) =>
y' = (x2 + x + 1)' + (x5 + 7)' = 2x1 + 1 + 0 + 5x4 + 0 = 5x4 + 2x + 1

If a function is a multiple of two functions the derivate is given by:

y = f(x).g(x) => y' = f'(x)g(x) + f(x)g'(x)

If f(x) = C(C is a contstant), and y = f(x)g(x)
y = Cg(x) y'=C'.g(x) + C.g'(x) = 0 + C.g'(x) = C.g'(x)

y = Cf(x) => y' = C.f'(x)

There are examples of the following formulas in the task section.

y =
 f(x) g(x)
y' =
 f'(x)g(x) - f(x)g'(x) g2(x)
y = ln x => y' = 1/x
y = ex => y' = ex
y = sin x => y' = cos x
y = cos x => y' = -sin x
y = tan x => y' = 1/cos2x
y = cot x => y' = -1/sin2x
y = arcsin x  =>  y' =
 1 √1 - x⋅x
y = arccos x  =>  y' =
 -1 √1 - x⋅x
y = arctan x  =>  y' =
 1 1 + x2
y = arccot x  =>  y' =
 -1 1 + x2

When a function is a function of function: u = u(x)

y = f(u) => y' = f'(u).u'

Example: let's given y = sin(x2)
Here u = x2, f(u) = sin(u), the derivatives are f'(u) = cos(u), u' = 2x
y' = (sin(u))'⋅u' = cos(x2)⋅2x = 2⋅x⋅cos(x2)

#### Problems involving derivatives

1) f(x) = 10x + 4y, What is the first derivative f'(x) = ?
Solution: We can use the formula for the derivate of function that is the sum of functions
f(x) = f1(x) + f2(x), f1(x) = 10x, f2(x) = 4y for the function f2(x) = 4y, y is a constant because the argument of f2(x) is x so f'2(x) = (4y)' = 0. Therefore, the derivative function of f(x) is: f'(x) = 10 + 0 = 10.

2) Calculate the derivative of f(x) =
 x10 4.15 + cosx

Solution: We have two functions h(x) = x10 and g(x) = 4.15 + cos x
the function f(x) is h(x) divided by g(x). h'(x) = 10x9 g'(x) = 0 - sin x = -sin x

f'(x) =
 h'(x).g(x) - h(x).g'(x) (g(x))2
f'(x) =
 10x9(4.15 + cos x) - x10(-sin x) (4.15 + cosx)2
=
 x10sin x + 10(60 + cos x)x9 (60 + cosx)2

3) f(x) = ln(sinx). what is the derivative of the function f(x)?
Solution: To solve the task we have to use the last formula. As we can see f(x) is a function of function of function f(x) = h(g(x)) where h = ln, and g = sin x

f'(x) =
 1 g(x)
g'(x) =
 1 sin x
cos x =
 cos x sin x

#### More about derivatives in the maths forum

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