find the derivative of
y=2cotx+sec3x+cscX
find d^2/dx^2 for 1+y=x+xy
The derivatives of these trig functions should be right there in your Calculus text.
for the second of your questions, I will use y' for the first derivative and y" for the second derivative i.e. y'= dy/dx
1+y=x+xy so
y' = 1 + y + xy'
y'(1-x) = 1+y
y' = (1+y)/(1-x)
y" = [(1-x)y' - (1+y)(-1)]/((1-x)^2)
=[(1+y) + (1+y)]/((1-x)^2) after I replaced y'
=(2+2y)/((1-x)^2)
To find the derivative of the given function y = 2cot(x) + sec(3x) + csc(x), we can apply the rules of differentiation. Let's break down the problem step by step:
Step 1: Differentiate each term in the function separately.
To differentiate cot(x), we can use the chain rule. The derivative of cot(x) is -(csc(x))^2.
To differentiate sec(3x), we use the chain rule as well. The derivative of sec(3x) is 3sec(3x)tan(3x).
To differentiate csc(x), we again use the chain rule. The derivative of csc(x) is -csc(x)cot(x).
Step 2: Combine the derivatives of each term.
Putting it all together, the derivative of y = 2cot(x) + sec(3x) + csc(x) is:
dy/dx = -(2csc^2(x)) + 3sec(3x)tan(3x) - csc(x)cot(x)
Now, to find d^2/dx^2 for the equation 1 + y = x + xy, we need to take the second derivative of y with respect to x.
Step 1: Differentiate the expression y = x + xy once.
Taking the derivative of x + xy with respect to x gives us:
dy/dx = 1 + y + x(dy/dx)
Step 2: Solve for dy/dx.
Rearrange the equation to solve for dy/dx:
dy/dx - x(dy/dx) = 1 + y
Factor out dy/dx:
(1 - x)dy/dx = 1 + y
Now, solve for dy/dx:
dy/dx = (1 + y) / (1 - x)
Step 3: Differentiate dy/dx with respect to x again.
Now, we take the derivative of dy/dx = (1 + y) / (1 - x) with respect to x:
d^2y/dx^2 = [(d(1 + y)/dx)(1 - x) - (1 + y)(d(1 - x)/dx)] / (1 - x)^2
Step 4: Simplify the expression.
Substitute the values of dy/dx and solve the expression:
d^2y/dx^2 = [(1 + y)(-1) - (1 - x)(dy/dx)] / (1 - x)^2
Simplify further:
d^2y/dx^2 = [-1 - y + (1 - x)((1 + y) / (1 - x))] / (1 - x)^2
d^2y/dx^2 = (-1 - y + 1 + y) / (1 - x)^2
d^2y/dx^2 = 0 / (1 - x)^2
d^2y/dx^2 = 0
Therefore, d^2/dx^2 for the equation 1 + y = x + xy is 0.