Question

If $y = {x^4} - 10$ and if x changes from 2 to 1.99, what is the approximate change in y? Also, find the changed value of y.

Answer 1

We write the initial value of $x$ is $2$ and the value of $x$ after the change $\Delta x$ i.e. $x + \Delta x$ as $1.99$, subtracting the original value from changed value we get the approximate change in x which can be written as $dx$. Using the method of differentiation, we differentiate $y$ with respect to $x$ and find the value of differentiation at the initial value of $x$. We find the value of $dy$ by multiplying the differentiation by $dx$ which gives us the change in y i.e. $\Delta y$. From the initial equation of $y$ we find the value of $y$ when $x$ is $2$ and add the change in $y$ to get the approximate value of $y$.

Formula used:

• Differentiation is given by $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$.
• $\Delta$ represents the approximate change in a quantity.

Complete Step-by-step Solution
We have $x = 2$and $x + \Delta x = 1.99$
So, we can find value of approximate change in $x$ by subtracting initial value of $x$ from new value of $x$ $\Rightarrow x + \Delta x - x = 1.99 - 2$
$\Rightarrow \Delta x = - 0.01$
We can write change in $x$ as $dx = \Delta x$
Now we have $y = {x^4} - 10$ … (1)
Differentiate equation (1) with respect to $x$ so we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({x^4} - 10)$
$\Rightarrow \dfrac{{dy}}{{dx}} = 4{x^3}$
We can write $dy = \dfrac{{dy}}{{dx}} \times dx$
$\Rightarrow dy = 4{x^3} \times dx$ … (2)
Now put the value of $x = 2$in equation (2) and $dx = \Delta x = - 0.01$
$\Rightarrow dy = 4{(2)^3} \times ( - 0.01)$
$\Rightarrow dy = 4 \times 8 \times ( - 0.01)$
$\Rightarrow dy = 32 \times ( - 0.01)$
When we change the value of the $x$ from 2 to 1.99 the change in value of $y$ is
$\Rightarrow dy = - 0.32$
We can write $dy = \Delta y$ as it represents change in value of $y$.
Now we calculate the original value of $y$ when $x$ is 2.
$\Rightarrow y = {x^4} - 10$
Substitute $x$ as $2$
$\Rightarrow y = {2^4} - 10$

$$\Rightarrow y = 16 - 10 \\\ \Rightarrow y = 6 \\\$$

Now we know change in y is $\Delta y = - 0.32$
So, New y after change in y will be given by $y + \Delta y$

$$\Rightarrow y + \Delta y = 6 + ( - 0.32) \\\ \Rightarrow y + \Delta y = 6 - 0.32 \\\ \Rightarrow y + \Delta y = 5.68 \\\$$

$y = 5.68$

$\therefore$ When the value of $x$ changes from 2 to 1.99 the changed value of $y$ is 5.68 and the approximate change in value of $y$ is $0.32$.

Note:
Students many times think that approximate change will always be positive which is wrong because the change will depend on initial and new value of the number, if the initial value of number is greater than new value, we will get approximate change in value as negative. Approximate change in a value tell us if we are adding value of subtracting value to initial value.
$\dfrac{{dy}}{{dx}}$ is the ratio of the change in value of $y$ when the change in the value of $x$ to the change in the value of $x$.