Find the area of a polygon with the given vertices A(2,2) B(5,2) C(3,-1) D(0,-1) Number 21

Use logarithmic differentiation to find dy/dx

liq [111]

Answer:

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx = sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx = 2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx = [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx = (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

7 0

3 years ago

67. The line contains the point (4,0) and is parallel to the line defined by 3x = 2y.

olganol [36]

Answer:

$y=\frac{3}{2} x-6$

Step-by-step explanation:

Hi there!

What we need to know:

Linear equations are typically organized in slope-intercept form:

$y=mx+b$ where m is the slope of the line and b is the y-intercept (the value of y when the line crosses the y-axis)

Parallel lines will always have the same slope but different y-intercepts.

1) Determine the slope of the parallel line

Organize 3x = 2y into slope-intercept form. Why? So we can easily identify the slope, m.

$3x = 2y$

Switch the sides

$2y=3x$

Divide both sides by 2 to isolate y

$\frac{2y}{2} = \frac{3}{2} x\\y=\frac{3}{2} x$

Now that this equation is in slope-intercept form, we can easily identify that $\frac{3}{2}$ is in the place of m. Therefore, because parallel lines have the same slope, the parallel line we're solving for now will also have the slope $\frac{3}{2}$ . Plug this into $y=mx+b$ :