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 a type y = xⁿ the formula for the derivative is:

y = xⁿ => y' = nx^n-1

example: y = x³ y' = 3x^3-1 = 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 multiplication of two functions the derivate formula is:

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 = e^x => y' = e^x

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 - x.x

y = arccos x

y' =

-1

√1 - x.x

y = arctan x

y' =

1 + x²

y = arccot x

y' =

-1

1 + x²

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

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

example: let's have function y = sin(x²)
in this situation 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)

Derivative Problems

1) f(x) = 10x + 4y, what will be the first derivative f'(x) = ?
ANSWER: We can use the formula for the derivate of function that is 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. So the derivative function of f(x) is: f'(x) = 10 + 0 = 10.

2) Calculate the derivative of f(x) =

x¹⁰

4.15 + cosx

ANSWER: 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)?
ANSWER: 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) =

g(x)

g'(x)

sin x

cos x

sin x

More about derivatives in the maths forum

Forum registration

Edit
Create a new page
Send us math lessons, lectures, tests, to: