
Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).




Solve for d<em>y</em>/d<em>x</em> :



If <em>y</em> ≠ 0, we can write

At the point (1, 1), the derivative is

X>_ 0
If you don’t understand that then x is greater than or equal to 0
we have the following:

solving for b:

therefore, the answer is 16
order of operations
BEMDAS:
B - <em>Brackets</em>
E - <em>Exponents</em>
D - <em>Division</em>
M - <em>Multiplication</em>
A - <em>Addition</em>
S - <em>Subtraction</em>
73 • 7 - 5 = 511 - 5 = 506
What comes next is 0.75
The pattern rule is you add 0.15