5. Find dy/dx of
of the following
the following functions
y²= 4ax
​

A general rule for functions of the general form:

y = A*x^n

is:

dy/dx =  n*A*x^(n - 1)

Then, for the function:

y^2 = 4*a*x

Then we can rewrite this as:

y = (4*a*x)^(1/2) = √(4*a)*x^(1/2)

Then we have:

A = √(4*a)

n = 1/2

Now we can use the above rule to get

dy/dx = (1/2)*√(4*a)*x^(1/2 - 1)

dy/dx = (1/2)*√(4*a)*x^(-1/2)

dy/dx = (1/2)*√(4*a/x)

dy/dx = √(1/4)*√(4*a/x)

dy/dx = √( (1/4)*4*a/x)

dy/dx = √(a/x)

3/6 chance

step-by-step explanation:

half of the nubers are odd numbers (1,3,5) which in that case if you would be told to simplify it would be 1/2

where is the diagram? shown below where? ?

step-by-step explanation: