Question

Find $\Delta y\,and\,\,dy$ for the following functions for the values of $x\,\,and\,\,\Delta x$ which are shown against each of the functions. (i) $y = {x^2} + 3x + 6,\,x = 10\,\,and\,\,\Delta x = 0.01$

Answer 1

To solve this problem , first consider $y = f(x)$ is a differentiable function of $x$ then $f'(x)dx$ is called the differential of $f$. Then,$dy = f'x\,fx$

Complete step by step solution:
$\Delta y = f(x + \Delta x) - fx$
$\Delta y = {(x + \Delta x)^2} + 3(x + \Delta x) + 6 - ({x^2} + 3x + 6)$
By using the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab,$ we will get
$\Delta y = {x^2} + {(\Delta x)^2} + 2x.\Delta x + 3x + 3\Delta x + 6 - {x^2} - 3x - 6$
$\Delta y = {(\Delta x)^2} + 2x.\Delta x + 3x$
Now, we put $n = 10$and $\Delta x = 0.01$in the above equation, we have
$\Delta y = {(0.01)^2} + 2 \times 10(0.01) + 3(0.01)$
$\Delta y = 0.01 \times 0.01 + 20 \times 0.01 + 0.03$
$\Delta y = 0.00001 + 0.2 + 0.03$
$\Delta y = 0.230$
Now, $y = {x^3} + 3x + 6$
Differentiate with respect to $x$, we will get
$dy = f'(x)dx$
Here $f(x) = {x^2} + 3x + 6$
Differentiate with respect to $x$
Then,$f(x) = 2x + 3$.
$dy = (2x + 3)(0.01) \\\ dy = (2 \times 10 + 3)(0.01) \\\ dy = (20 + 3)(0.01) \\\$
$dy = 23 \times 0.01$
$dy = 0.23$
Hence, $\Delta y = 0.2301$ and $dy = 0.23$

Note: If each input value leads to only one output value, the relationship is a function. If any input value leads to two or more outputs, the relationship is not a function.