Question

If $y = 2{x^3} - 2{x^2} + 3x - 5$, then for $x = 2$ and $\Delta x = 0.1$, value of $\Delta y$ is
1. $2.002$
2. $1.9$
3. $0$
4. $0.9$

Answer 1

Hint: In this question, we are given the function $y = 2{x^3} - 2{x^2} + 3x - 5$ and we have to find the value of $\Delta y$ where the value of $x = 2$ and $\Delta x = 0.1$. First step is to differentiate the function with respect to $x$ and change the terms into $\Delta x = dx$ and $\Delta y = dy$. Solve further by putting the given values.

Formula Used:
Differentiation formula – We find the instantaneous rate of change of one quantity with respect to another, the process is called differentiation. We called it differentials in calculus because of this minor difference.
$\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$

Complete step by step Solution:
Given that,
$y = 2{x^3} - 2{x^2} + 3x - 5$
Differentiate the above equation with respect to $x$,
\[\dfrac{{dy}}{{dx}} = 6{x^2} - 4x + 3\]
As we know that, $\Delta x = dx$ and $\Delta y = dy$ Change them in the above equation
\[\dfrac{{\Delta y}}{{\Delta x}} = 6{x^2} - 4x + 3\]
Cross multiplying both sides,
\[\Delta y = 6{x^2}\left( {\Delta x} \right) - 4x\left( {\Delta x} \right) + 3\Delta x\]
Now, the value of $x,\Delta x$ are given in the question i.e., $x = 2$ and $\Delta x = 0.1$,
\[\Delta y = 6{\left( 2 \right)^2}\left( {0.1} \right) - 4\left( 2 \right)\left( {0.1} \right) + 3\left( {0.1} \right)\]
\[\Delta y = 2.4 - 0.8 + 0.3\]
\[\Delta y = 0.9\]
Hence, Option (4) is the correct answer i.e., $0.9$.

Hence, the correct option is 4.

Note: The key concept involved in solving this problem is a good knowledge of differentiation. Students must remember that differentiation is a technique for determining a function's derivative. Differentiation is a mathematical process that determines the instantaneous rate of change of a function based on one of its variables. The most common example is velocity, which is the rate of change of displacement with respect to time. To solve such questions, one should always know that $\Delta x,\Delta y$is nothing but another form of $dx,dy$.