Question

determine the slope of the tangent to the curve at the point y=(3x)(sinx) at the point with x-coordinate pi/2 ​

Answer 1

Answer:

Step-by-step explanation:

The slope of the tangent to a curve is the derivative of the curve. We need to find the derivative of the function and then evaluate the derivative at that given x value. The derivative is found using the product rule:

$y'=f(x)g'(x)+f'(x)g(x)$

Let's call 3x our f(x) and sin(x) our g(x). Filling in the formula for the derivative using the product rule looks like this:

$y'=3x(cosx)+3sinx$

That gives us the derivative, which is the slope formula that can be used at ANY x value anywhere on the curve to find the slope of the line tangent to the curve at that x value. If we want to find the slope of the tangent line to the curve at x = pi/2, we evaluate the slope formula at x = pi/2 (remember that y' is the same exact thing as the slope):

$y'(slope)=3(\frac{\pi}{2})(cos\frac{\pi}{2})+3(sin\frac{\pi}{2})$

From the unit circle (or experience, since you're in advaced math), we know that the cosine of pi/2 is 0 and that the sin of pi/2 = 1:

$y'(slope)=3(\frac{\pi}{2})(0)+3(1)$ simplifies to

y' (slope) = 3

That means that the slope of the line tangent to the curve at the point x = pi/2 is 3.