Question

Find $ \vartriangle f $ and $ df $ for the function f for the indicated values of x, $ \vartriangle x $ and compare $ f(x) = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5 $

Answer 1

Hint: First, differentiation can be defined as the derivative of independent variables value and can be used to calculate feature independent variable per unit modification.
Let $ y = f(x) $ be the given function of x, the differentiation gives $ \dfrac{{dy}}{{dx}} $ (with y-respect to)
The most popular power rule for differentiation is $ \dfrac{d}{{dx}}{x^n} = n{x^{n - 1}} $

Complete step by step answer:
From the given that we have, $ f(x) = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5 $
Since f(x) is the function of y and both are equally as in the domain and codomain for x is the domain and y are the codomains.
Hence take $ f(x) = y $ , rewrite the given problem we get, $ y = {x^3} - 2{x^2};x = 2,\vartriangle x = dx = 0.5 $
By the differentiation rule, now we are going to differentiate the function y.
Thus, we get, $ y = {x^3} - 2{x^2} \Rightarrow \dfrac{{dy}}{{dx}} = 3{x^2} - 4x $
Now equating the denominator dx into the right-hand side values we get, $ \dfrac{{dy}}{{dx}} = 3{x^2} - 4x \Rightarrow dy = (3{x^2} - 4x)dx $
From this we have some values for the function that $ x = 2,\vartriangle x = dx = 0.5 $ where the value x is given as two and differentiation is given as zero points five.
subsisting the values in the converted equation we get, $ dy = (3{x^2} - 4x)dx \Rightarrow dy = (3{(2)^2} - 4(2))[0.5] $
Where $ x = 2,\vartriangle x = dx = 0.5 $ .
Further solving the equation, we get, $ dy = (3{(2)^2} - 4(2))[0.5] \Rightarrow dy = (12 - 8)[0.5] \Rightarrow 2 $ where $ 4 \times 0.5 = 2 $ .
Hence, we get the differentiation value of $ df = 2 $
Now we are going to find the del value of the same function given Which is $ f(x) = {x^3} - 2{x^2} $
Now converting the function into del function thus we get, $ \vartriangle f = f(x + \vartriangle x) - f(x) $
Since from the given that we have $ x = 2,\vartriangle x = dx = 0.5 $ , applying this we get, $ \vartriangle f = f(2 + 0.5) - f(2) $
First, we will find the term one in the del value, $ f(2 + 0.5) $ , now convert this value into the $ f(x) = {x^3} - 2{x^2} $ original function we get, $ f(2 + 0.5) = f(2.5) \Rightarrow {(2.5)^3} - 2{(2.5)^2} $ and further solving this we get, $ f(2.5) = 3.125 $
For the second term, we have $ f(2) = {(2)^3} - 2{(2)^2} \Rightarrow 0 $
Hence, we get, $ \vartriangle f = f(2 + 0.5) - f(2) \Rightarrow 3.125 $
Therefore $ df = 2 $ and $ \vartriangle f = 3.125 $ are the values of the given equation.

Note: Since the formation of the del x is the derivation of the given function represented as $ \vartriangle f = f(x + \vartriangle x) - f(x) $ .
Derivative of any constant multiplied with the function f: $ \dfrac{d}{{dx}}(a.y) = a{y^1} $ (the constant values in the differentiation of integration will be not changed in any format)
The chain rule of the two functions is representing as $ \dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}} $