help find dy/dx if y=cot(x+y)
working plz
y = cot(x+y)
y' = -csc^2(x+y) (1+y')
y' = -csc^2(x+y) - csc^2(x+y) y'
y' (1+csc^2(x+y)) = -csc^2(x+y)
y' = -csc^2(x+y)/(1+csc^2(x+y))
But, csc^2 = 1+cot^2, so
y' = -(1+y^2)/(1+1+y^2)
= -(1+y^2)/(2+y^2)
or, you can say
1/y = tan(x+y)
arctan(1/y) = x+y
1/(1 + 1/y^2))(-1/y^2)y' = 1+y'
-1/(1+y^2) y' = 1+y'
y' (1 + 1/(1+y^2)) = -1
y' (2+y^2)/(1+y^2) = -1
y' = -(1+y^2)/(2+y^2)
To find the derivative of y = cot(x + y), we will use implicit differentiation.
Step 1: Start by taking the derivative with respect to x of both sides of the equation.
d/dx (y) = d/dx (cot(x + y))
Step 2: For the left side, we need to use the chain rule since y is a function of x.
dy/dx = d/dx(cot(x + y))
Step 3: For the right side, we also need to use the chain rule. The derivative of cot(u) is -(csc^2(u)), and we treat (x + y) as a single variable.
dy/dx = -csc^2(x + y) * (1 + dy/dx)
Step 4: Multiply out the -csc^2(x + y) term.
dy/dx = -csc^2(x + y) - csc^2(x + y) * dy/dx
Step 5: Rearrange the equation to solve for dy/dx.
dy/dx + csc^2(x + y) * dy/dx = -csc^2(x + y)
Step 6: Factor out dy/dx on the left side.
dy/dx(1 + csc^2(x + y)) = -csc^2(x + y)
Step 7: Divide both sides by (1 + csc^2(x + y)).
dy/dx = -csc^2(x + y) / (1 + csc^2(x + y))
Therefore, dy/dx = -csc^2(x + y) / (1 + csc^2(x + y)).
Please let me know if anything is unclear or if you need further assistance!
To find dy/dx, we need to differentiate the equation y = cot(x + y) with respect to x using the chain rule.
First, let's rewrite the equation in terms of sine and cosine functions:
y = cot(x + y)
y = cos(x + y)/sin(x + y)
Now, differentiate both sides of the equation with respect to x using the chain rule. Recall that the derivative of cot(u) is -cosec^2(u) * du.
dy/dx = [d/dx (cos(x + y))/sin(x + y)] - [cos(x + y)/sin(x + y)] * [d/dx (sin(x + y))]
Next, let's differentiate the individual terms:
For the first term, d/dx (cos(x + y)), we use the chain rule:
d/dx (cos(x + y)) = -sin(x + y) * [d/dx (x + y)]
= -sin(x + y) * (1 + dy/dx)
For the second term, d/dx (sin(x + y)), we also use the chain rule:
d/dx (sin(x + y)) = cos(x + y) * [d/dx (x + y)]
= cos(x + y) * (1 + dy/dx)
Substituting the derivatives back into the equation, we have:
dy/dx = [-sin(x + y) * (1 + dy/dx)]/sin(x + y) - [cos(x + y)/sin(x + y)] * [cos(x + y) * (1 + dy/dx)]
Simplifying, we get:
dy/dx = [-sin(x + y) - sin^2(x + y) * dy/dx]/sin(x + y) - cos^2(x + y) * (1 + dy/dx)/sin(x + y)
To solve for dy/dx, we can isolate the terms with dy/dx on one side:
dy/dx + cos^2(x + y) * dy/dx = -sin(x + y) - sin^2(x + y) * dy/dx - cos^2(x + y)
Taking out dy/dx as a common factor:
dy/dx (1 + cos^2(x + y) + sin^2(x + y)) = -sin(x + y) - cos^2(x + y)
Since cos^2(x + y) + sin^2(x + y) = 1, we have:
dy/dx = [-sin(x + y) - cos^2(x + y)] / [1 + cos^2(x + y) + sin^2(x + y)]
Simplifying further, we get:
dy/dx = -sin(x + y) - cos^2(x + y)