Find the slope of the tangent line to the curve below at the point (2,7)

Question

Find the slope of the tangent line to the curve below at the point (2,7)

√(3x+2y) + √(3xy) = 10.952876653407

Everything in the parentheses is under the square root.

Leave your answer in a negative decimal form.

What is the slope?

Answer 1

This is very much like the other problem I did for you except you left off an important statement in the other problem" Leave your answer in a negative decimal form." So here goes:

Write using a fractional exponent:

(3x + 2y)^1/2 + (3xy)^1/2 = 10.952876653407

Take the derivative of both sides, implicitly, with respect to "x" using the power rule, the chain rule, and the product rule:

(1/2)((3x + 2y)^-1/2(3 + 2y') + (1/2)(3xy)^-1/2(3y + 3xy') = 0

Solve for y':

(3/2)(3x + 2y)^-1/2 + y'(3x + 2y)^-1/2 + (3y/2)(3xy)^-1/2 + (3xy'/2)(3xy)^-1/2 = 0

y'(3x + 2y)^-1/2 + (3xy'/2)(3xy)^-1/2 = - (3/2)(3x + 2y)^-1/2 - (3y/2)(3xy)^-1/2

y'[(3x + 2y)^-1/2 + (3x/2)(3xy)^-1/2] = - (3/2)(3x + 2y)^-1/2 - (3y/2)(3xy)^-1/2

y' = [-(3/2)(3x + 2y)^-1/2 - (3y/2)(3xy)^-1/2]/[(3x + 2y)^-1/2 + (3x/2)(3xy)^-1/2]

Plug in x = 2 and y = 7:

y' = -2.848575935

You will need to decide how to round if desired.

Answer 2

Hello,

This is implicit differentiation.

(3x+2y)^(1/2)+(3xy)^(1/2)=10.952...

If we take derivative, then

1/2 * (3x+2y)^(-1/2) * (3+2(dy/dx)) + 1/2 * (3xy)^(-1/2) * (3y+3x(dy/dx)) = 0

[d/dx(3xy) = 3y + 3x (dy/dx) <- chain rule ]

Now we are looking for a slope at (2,7)

1/2 * (3*2+2*7)^(-1/2) * (3+2(dydx))+ 1/2*(3*2*7)^(-1/2)*(3*7+3*2(dy/dx))=0

1/2*(20)^(-1/2)*(3+2dy/dx)+1/2(42)^(-1/2)(21+6dy/dx)=0

[20^(-1/2)=0.2236067977, 42^(-1/2)=0.15430335]

0.1118033989*3 + 0.1118033989*2*dy/dx+0.07715167498*21+0.07715167498*6*dy/dx=0

( 0.1118033989*2+0.07715167498*6)dydx=-0.1118033989*3-0.07715167498*21

0.6865168477*dy/dx=-1.955595371

dy/dx = -2.848575935 at (2,7)