# Use implicit differentiation to find an equation of the tangent line to the graph of the equation at the given point. arcsin(4x)+arcsin(3y)= pi/2, (`sqrt(2)/8` , `sqrt(2)/6` )

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To find the equation of a tangent line, we need the slope and the point it goes through.

This is the result after implicit differentiation.

`4/(sqrt(1-16x^2)) + 3/(sqrt(1-9y^2)) (dy)/(dx) = 0`

In order to find the slope at that point, we will solve for dy/dx and plug in our point for the y value.

`3/(sqrt(1-9y^2)) (dy)/(dx) = -4/(sqrt(1-16x^2))`

`(dy)/(dx) = -4/(sqrt(1-16x^2)) * sqrt(1-9y^2)/3`

`(dy)/(dx) = -4/(sqrt(1-16(sqrt(2)/8)^2)) * sqrt(1-9(sqrt(2)/6)^2)/3`

`dy/dx = -4/sqrt2 * sqrt2/3`

`dy/dx = -4/3`

Our slope is -4/3, our points are `(sqrt2/8, sqrt2/6)`

Plug that into point slope form

`y - sqrt2/6 = -4/3 (x - sqrt2/8)`

This is the equation of the tangent line at that point.

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so in order to find the y=

would i need to move the `sqrt(2)/6` to the other side? and would i need to distribute the -4/3 into the x and `sqrt(2)/8` ?

Yes, if you want to put it into y=mx + b form you move the root(2)-6 to the right side, and then distribute and combine the constant terms.

the implicit differentiation of arcsin(4x)+arcsin(3y)= pi/2 is shown in figure ,

we get: y' = -(4 sqrt(1-9 y^2))/(3 sqrt(1-16 x^2)

find the value of y' at point ( , )

then y' = -(4 sqrt(1-9( )^2))/(3 sqrt(1-16()^2)

y' = -(4 sqrt(1 - 1/2))/(3 sqrt(1 - 1/2)

y' = -(4 sqrt(1/2))/(3 sqrt(1/2)

y' = -4/3therefore, the equation if tangent:

y - y1 = y'(x - x1)

y - = -4/3( x - )

y - = -4/3 x +

y + 4/3 x = sqrt(2)/3