The first derivative of the function f(x) = x² - 5 is equal to f'(x) = 2 · x.
<h3>How to find the derivative of a quadratic equation by definition of derivative</h3>
In this question we have a quadratic function, in which we must make use of the definition of derivative to find the expression of its first derivative. Then, the procedure is shown below:
f(x) = x² - 5 Given
f' = [(x + h)² - 5 - x² + 5] / h Definition of derivative
(x² + 2 · x · h + h² - 5 - x² + 5) / h Perfect square trinomial
(2 · x · h + h²) / h Associative, commutative and modulative properties / Existence of additive inverse
2 · x + h Distributive, commutative and associative properties / Definition of division / Existence of multiplicative inverse
2 · x h = 0 / Result
The first derivative of the function f(x) = x² - 5 is equal to f'(x) = 2 · x.
To learn more on derivatives: brainly.com/question/25324584
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Answer:
9
Step-by-step explanation:
You do parenthesis first so 1+2=3
Then you do division 6 divided by 2=3
Then you will multiply 3 by 3 so 9
This is known as pemdas
P=parenthesis
E=exponents
MD=multiplication and division you do it left to right so if division is in front of multiplication then you do division
As=Addition and subtraction is also left to right
Answer: 36000
Step-by-step explanation:
try looking it up lol
Answer:
n = 10
Step-by-step explanation:
given m varies directly as n then the equation relating them is
m = kn ← k is the constant of variation
to find k use the condition m = 6 when n = 5
6 = 5k ( divide both sides by 5 )
k = 
= 1.2
m = 1.2n ← equation of variation
when m = 12 , then
12 = 1.2n ( divide both sides by 1.2 )
10 = n
