what is the slope of a line tangent to y=sin^2(5x+pi) at x=pi/4?
Take the derivative. Remember to apply the chain rule!
y'=[2sin(5x+pi)](5)
plug in pi/4
y'=[2sin(5x+pi)]*(cos(5x+pi))*5
look at example 3:
http://web.mit.edu/wwmath/calculus/differentiation/chain.html
thank you
To find the slope of a line tangent to a curve at a specific point, you can take the derivative of the curve and then substitute the x-coordinate of the point of interest into the derivative equation.
In this case, we have the curve y = sin^2(5x+pi).
Step 1: Take the derivative of the curve.
To differentiate y = sin^2(5x+pi), we can use the chain rule. The derivative of sin^2(u) with respect to u is 2sin(u)cos(u). Applying the chain rule, we have:
dy/dx = d/dx[sin^2(5x+pi)]
= 2sin(5x+pi)cos(5x+pi) * d/dx[5x+pi]
= 2sin(5x+pi)cos(5x+pi) * 5
= 10sin(5x+pi)cos(5x+pi)
Step 2: Substitute the x-coordinate of the point of interest.
The x-coordinate of the point of interest is x = pi/4. We need to substitute this value into the derivative equation we obtained in Step 1:
dy/dx = 10sin(5(pi/4)+pi)cos(5(pi/4)+pi)
= 10sin(5(pi/4)+pi)cos(5(pi/4)+pi)
Step 3: Simplify the expression.
To simplify, we can use the trigonometric identities:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos(a + b) = cos(a)cos(b) - sin(a)sin(b)
10sin(5(pi/4)+pi)cos(5(pi/4)+pi)
= 10[sin(5(pi/4))cos(pi) + cos(5(pi/4))sin(pi)]
= 10[sin(5(pi/4))(-1) + cos(5(pi/4))(0)]
= 10[-sin(5(pi/4))]
= -10sin(5(pi/4))
So, the slope of the line tangent to the curve y = sin^2(5x+pi) at x = pi/4 is -10sin(5(pi/4)).