Answer:
Step-by-step explanation: Derivatives are used to find the rate of changes of a quantity with respect to the other quantity. By using the application of derivatives we can find the approximate change in one quantity with respect to the change in the other quantity. Assume we have a function y = f(x), which is defined in the interval [a, a+h], then the average rate of change in the function in the given interval is
(f(a + h)-f(a))/h
Now using the definition of derivative, we can write
f
'
(
a
)
=
lim
h
→
0
f
(
a
+
h
)
−
f
(
a
)
h
which is also the instantaneous rate of change of the function f(x) at a.
Now, for a very small value of h, we can write
f'(a) ≈ (f(a+h) − f(a))/h
or
f(a+h) ≈ f(a) + f'(a)h
This means, if we want to find the small change in a function, we just have to find the derivative of the function at the given point, and using the given equation we can calculate the change. Hence the derivative gives the instantaneous rate of change of a function within the given limits and can be used to find the estimated change in the function f(x) for the small change in the other variable(x).
Approximation Value
Derivative of a function can be used to find the linear approximation of a function at a given value. The linear approximation method was given by Newton and he suggested finding the value of the function at the given point and then finding the equation of the tangent line to find the approximately close value to the function. The equation of the function of the tangent is
L(x) = f(a) + f'(a)(x−a)
The tangent will be a very good approximation to the function's graph and will give the closest value of the function. Let us understand this with an example, we can estimate the value of √9.1 using the linear approximation. Here we have the function: f(x) = y = √x. We will find the value of √9 and using linear approximation, we will find the value of √9.1.
We have f(x) = √x, then f'(x) = 1/(2√x)
Putting a = 9 in L(x) = f(a) + f'(a)(x−a), we get,
L(x) = f(9) + f'(9)(9.1−9)
L(x) = 3 + (1/6)0.1
L(x) ≈ 3.0167.
This value is very close to the actual value of √(9.1)
Hence by using derivatives, we can find the linear approximation of function to get the value near to the function.