# Find the slope of the line tangent to the curve `(x+2y)^(1/2)+(3xy)^(1/2)=17.7` at the point (8,7) .(1x+2y)^1/2+(3xy)^1/2=17.7 at the points (8,7)

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### 2 Answers

Note that the slope of a line which is tangent to a point in a curve is equal to y'.

So to solve for the above problem, take the derivative of both sides of the given equation. Use implicit differentiation since x and y variables are on the same side of the equation.

`(x+2y)^(1/2) + (3xy)^(1/2)=17.7`

`d/(dx)((x+2y)^(1/2) + (3xy)^(1/2))=d/(dx)17.7`

`d/(dx)(x+2y)^(1/2) + d/(dx)(3xy)^(1/2)=d/(dx)17.7`

Then, let's take the derivative of each term separately.

For the first term, use power formula of derivative which is `(u^n)'=n*u^(n-1)*u'` .

>> `d/(dx) (x+2y)^(1/2)=1/2(x+2y)^(-1/2)*(x+2y)'`

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

Express the exponent as positive.

`=(1+2y')/(2(x+2y)^(1/2))`

For the second term, use the power formula of derivatives, too.

>> `d/(dx)(3xy)^(1/2)=1/2(3xy)^(-1/2)*(3xy)'`

To determine (3xy)', use the product rule which is`(u*v)'=vu'+uv'`.

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

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

Also, express the exponent as positive.

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

And at the right side, note that the derivative of a constant is zero. So,

>> `d/(dx)17.7=0`

Thus, taking the derivative of both sides of the equation result to:

`d/(dx)(x+2y)^(1/2) + d/(dx)(3xy)^(1/2)=d/(dx)17.7`

From the given point (8,7), substitute x=8 and y=7.

`(1+2y')/(2(8+2*7)^(1/2))+ (3*7+3*8y')/(2(3*8*7)^(1/2))=0`

`(1+2y')/(2*22^(1/2))+(21+24y')/(2*(168)^(1/2))=0`

`(1+2y')/(2*22^(1/2))+(21+24y')/(4*42^(1/2))=0`

To cancel the denominators, multiply both sides by the LCD.

`(4*22^(1/2)*42^(1/2))[(1+2y')/(2*22^(1/2))+(21+24y')/(4*42^(1/2))]=0`

`2*42^(1/2)(1+2y')+22^(1/2)(21+24y')=0`

`2*42^(1/2)+4*42^(1/2)y'+21*22^(1/2)+24*22^(1/2)y'=0`

Move the terms without y' to the right side.

`4*42^(1/2)y'+24*22^(1/2)y'=-2*42^(1/2)-21*22^(1/2)`

Factor out y' at the left side.

`y'(4*42^(1/2)+24*22^(1/2))=-2*42^(1/2)-24*22^(1/2)`

Then, isolate y'.

`y'=-(2*42^(1/2)+21*22^(1/2))/(4*42^(1/2)+24*22^(1/2))`

`y'=-0.805`

**Hence, the slope of the tangent line is -0.805.**

The slope of a tangent to any curve is given by the first derivative.

`(x+2y)^(1/2)+(3xy)^(1/2) = 17.7`

Derivate both sides with respect to x.

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

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

The slope of curve is given by y' at point (8,7).

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

`1/(2*(8+14)^(1/2))+(2y')/(2*(8+14)^(1/2))+(21)/(2*(3*56)^(1/2))+(24y')/(2*(3*56)^(1/2)) = 0`

`1/(2*(22)^(1/2))+(y')/((22)^(1/2))+(21)/(2*(168)^(1/2))+(12y')/((168)^(1/2)) = 0`

`0.1066+0.213y'+0.81+0.926y' = 0`

`1.139y' = -0.9166`

`y' = -0.8047`

*So the slope of the tangent is -0.8047.*

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