Question 1191910
For the following equations connecting x and y, if the rate of change of y is 4 units s^-1,<pre>
That says that
{{{dy/dt=4}}}{{{units/second}}}

You didn't put in any parentheses, so I can't tell whether you meant
{{{y=sqrt(2x+7)}}} or {{{y=sqrt(2x)+7}}}. I'll assume

{{{y=sqrt(2x+7)}}}

Write the square root as the {{{1/2}}} power:

{{{matrix(2,1,"",y=(2x+7)^(1/2))}}} 

Take the derivative with respect to t

{{{matrix(2,1,"",dy/dt=expr(1/2)(2x+7)^(-1/2)(2*expr(dx/dt)))}}}

Substitute 4 for {{{dy/dt}}} and simplify

{{{matrix(2,1,"",4=expr(1/cross(2))(2x+7)^(-1/2)(cross(2)*expr(dx/dt)))}}}

{{{matrix(2,1,"",4=(2x+7)^(-1/2)(expr(dx/dt)))}}}

Bring the negative exponent to the denominator as a positive exponent

{{{matrix(2,1,"",4=(1/(2x+7)^("1/2"))(expr(dx/dt)))}}}

Change the {{{1/2}}} power to a square root

{{{4=(1/sqrt(2x+7))expr(dx/dt)}}}</pre>

find the
rate of change of x at the given instant: y = 3<pre>

Go back to {{{y=sqrt(2x+7)}}}

Substitute y=3 and solve for x:

{{{3=sqrt(2x+7)}}}
Square both sides:
{{{9=2x+7}}}
Solve for x

Substitute that value of x in

{{{4=(1/sqrt(2x+7))expr(dx/dt)}}}

and solve for {{{dx/dt}}}.

Edwin</pre>