Question

matched the variable we are differentiating with respect to, we simply write, 2x. In the example below, the variables do not necessarily match. Find the derivative \"with respect to x\" of $x^2 + y^2 = 2$ The derivative of the $x^2$ is still 2x. However, the derivative of the $y^2$ would be 2y(dy/dx) since the variables do not match. In this exercise, CHOOSE ONE of the problems below and find the derivative with respect to x. Solve the expression for dy/dx. Then find the value of the derivative at the indicated point. Finally, state the equation of the tangent line at the indicated point. Recall that the equation of a line can be written using the equation $y - y_1 = m(x - x_1)$. The value of m in your problem will be the slope you attain from your work. $\sin(xy) = \frac{\sqrt{2}}{2}$ $\left(\frac{\pi}{2}, \frac{1}{2}\right)$ $x^2 + 3y^3 = 25$ $(1, 2)$ $e^x + e^y = 2$ $(0, 0)$ $e^{xy} = 1$ $x^2 = \cos y$ $x^2y^2 = 16$ $(1, 0)$ $(1, 0)$ $(2, 2)$ *******You may want to consider finishing all of the material from this unit prior to completing this assignment. Some of the problems in the chart involve the derivative of

          matched the variable we are differentiating with respect to, we simply write, 2x. In the
example below, the variables do not necessarily match.
Find the derivative \"with respect to x\" of $x^2 + y^2 = 2$
The derivative of the $x^2$ is still 2x. However, the derivative of the $y^2$ would be 2y(dy/dx) since the
variables do not match. In this exercise, CHOOSE ONE of the problems below and find the
derivative with respect to x. Solve the expression for dy/dx. Then find the value of the
derivative at the indicated point. Finally, state the equation of the tangent line at the
indicated point.
Recall that the equation of a line can be written using the equation $y - y_1 = m(x - x_1)$. The value of
m in your problem will be the slope you attain from your work.
$\sin(xy) = \frac{\sqrt{2}}{2}$
$\left(\frac{\pi}{2}, \frac{1}{2}\right)$
$x^2 + 3y^3 = 25$
$(1, 2)$
$e^x + e^y = 2$
$(0, 0)$
$e^{xy} = 1$
$x^2 = \cos y$
$x^2y^2 = 16$
$(1, 0)$
$(1, 0)$
$(2, 2)$
*******You may want to consider finishing all of the material from this unit prior to
completing this assignment. Some of the problems in the chart involve the derivative of

matched the variable we are differentiating with respect to, we simply write, 2x. In the
example below, the variables do not necessarily match.
Find the derivative ẅith respect to xöf x^2 + y^2 = 2
The derivative of the x^2 is still 2x. However, the derivative of the y^2 would be 2y(dy/dx) since the
variables do not match. In this exercise, CHOOSE ONE of the problems below and find the
derivative with respect to x. Solve the expression for dy/dx. Then find the value of the
derivative at the indicated point. Finally, state the equation of the tangent line at the
indicated point.
Recall that the equation of a line can be written using the equation y - y1 = m(x - x1). The value of
m in your problem will be the slope you attain from your work.
sin(xy) = (√(2))/(2)
((π)/(2), (1)/(2))
x^2 + 3y^3 = 25
(1, 2)
e^x + e^y = 2
(0, 0)
e^xy = 1
x^2 = cos y
x^2y^2 = 16
(1, 0)
(1, 0)
(2, 2)
*******You may want to consider finishing all of the material from this unit prior to
completing this assignment. Some of the problems in the chart involve the derivative of

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