Question:

Find dy/dx of ax+by^2=cosy

Answer:

Answer: Step 1: Rewrite the equation in terms of y: ax + by2 = cosy y = (ax + cosy)/b2

Step 2: Differentiate both sides with respect to x: dy/dx = (d(ax + cosy)/dx)/b2

Step 3: Apply the Chain Rule to the left side: dy/dx = (a + (dcosy/dx)*(-1))/b2

Step 4: Differentiate the right side with respect to x: dcosy/dx = -siny

Step 5: Substitute the result from Step 4 into Step 3: dy/dx = (a - siny)/b2

Question:

Find dy/dx of y=sin^−1(1−x^2/1+x^2),0<x<1

Answer:

Answer:

Step 1: Differentiate both sides with respect to x.

dy/dx = d/dx (sin^−1(1−x^2/1+x^2))

Step 2: Use the Chain Rule.

dy/dx = (d/dx (1−x^2/1+x^2)) * (d/dx (sin^−1(1−x^2/1+x^2)))

Step 3: Differentiate the numerator and denominator of the fraction.

dy/dx = [(−2x(1+x^2) - (1−x^2)(2x))/(1+x^2)^2] * (d/dx (sin^−1(1−x^2/1+x^2)))

Step 4: Use the formula for the derivative of inverse sine.

dy/dx = [(−2x(1+x^2) - (1−x^2)(2x))/(1+x^2)^2] * (1/(1-(1−x^2/1+x^2)^2)^(1/2))

Step 5: Simplify the expression.

dy/dx = [(-2x(1+x^2) - (1-x^2)(2x))/(1+x^2)^2] * [1/(1-(1-x^2/1+x^2)^2)^(1/2)]

Step 6: Final Answer.

dy/dx = -2x/(1+x^2)^(3/2)

Question:

Find dy/dx of sin^2x+cos^2y=1

Answer:

Answer:

Step 1: Take the derivative of each side of the equation with respect to x.

d/dx (sin2x + cos2y) = d/dx (1)

Step 2: Apply the chain rule on the left side of the equation.

2*sin(2x)cos(2x) + 0cos(2y) = 0

Step 3: Simplify the equation.

2*sin(2x)*cos(2x) = 0

Step 4: Factor out the 2*sin(2x).

2sin(2x)[cos(2x)] = 0

Step 5: Solve for the derivative.

2*sin(2x) = 0

sin(2x) = 0

2x = (2n + 1)π/2 (where n is an integer)

x = (n + 1/2)π (where n is an integer)

Question:

Find dy/dx of x^3+x^2y+xy^2+y^3=81

Answer:

Answer:

Step 1: Rewrite the equation as a function of y: y^3 + y^2(x+1) + y(x^2+1) + x^3 = 81

Step 2: Take the derivative of both sides with respect to x: 3x^2 + 2xy + y^2(1) + y(2x) = 0

Step 3: Simplify the equation: 3x^2 + 2xy + y^2 + 2xy = 0

Step 4: Factor out y from the equation: y(3x + 2x + 1) + y(2x + 1) = 0

Step 5: Divide both sides of the equation by (3x + 2x + 1): y + 2xy/(3x + 2x + 1) = 0

Step 6: Take the derivative of both sides with respect to x: dy/dx = -2y/(3x + 2x + 1)^2

Question:

Find dy/dx of y=tan^−1(3x−x^3/1−3x^2),−1/√3<x<1/√3

Answer:

Step 1: Rewrite y in terms of inverse tangent: y = tan^−1(3x−x^3/1−3x^2)

Step 2: Take the derivative of both sides with respect to x: dy/dx = (3 - 3x^2)/(1 - 3x^2)^2 * d/dx(3x - x^3)

Step 3: Take the derivative of the numerator: dy/dx = (3 - 3x^2)/(1 - 3x^2)^2 * (3 - 3x^2)

Step 4: Simplify: dy/dx = (3 - 3x^2)^2/(1 - 3x^2)^2

Question:

Find dy/dx of y=sin^−1(2x/1+x^2)

Answer:

Answer: Step 1: Rewrite the equation as y = arcsin(2x/(1+x^2)) Step 2: Take the derivative of both sides of the equation with respect to x, using the chain rule: dy/dx = (2/(1+x^2))(1) - (2x2x)/(1+x^2)^2 Step 3: Simplify the equation: dy/dx = 2/(1+x^2)^2 - 4x/(1+x^2)^2 Step 4: Simplify further: dy/dx = (2 - 4x)/(1+x^2)^2

Question:

Find dy/dx of 2x+3y=sinx

Answer:

Step 1: Rewrite the equation in terms of y: 2x + 3y = sin x 3y = sin x - 2x y = (sin x - 2x)/3

Step 2: Take the derivative of both sides with respect to x: dy/dx = -2/3 - (cos x)/3

Therefore, dy/dx = -2/3 - (cos x)/3

Question:

Find dy/dx of y=cos^−1(2x/1+x^2),−1<x<1

Answer:

Answer: Step 1: Differentiate y with respect to x using chain rule dy/dx = - (2/(1 + x^2)^2) * (1 + 2x^2)

Step 2: Simplify the expression dy/dx = - (2/(1 + x^2)^2) * (1 + 2x^2) = - (2/(1 + x^2)^2) * (1 + 2x^2) = - (2/(1 + x^2)^2) * (1 + 2x^2) = - (2/(1 + x^2)) * (2x) = - (4x)/(1 + x^2)

Question:

Find dy/dx of sin^2y+cosxy=k

Answer:

Answer: Step 1: Take the derivative of both sides of the equation with respect to x.

dy/dx = (2sin(y)cos(y) + cos(x)y)dx - 0

Step 2: Simplify the equation

dy/dx = (2sin(y)cos(y) + cos(x)y)dx

Step 3: Factor out dy

dy/dx = dy(2sin(y)cos(y) + cos(x)y)dx

Step 4: Divide both sides by (2sin(y)cos(y) + cos(x)y)

dy/dx = dy/ (2sin(y)cos(y) + cos(x)y)dx

Therefore, the answer is dy/dx = dy/ (2sin(y)cos(y) + cos(x)y)dx

Question:

Find dy/dx of y=sin^−1(2x√1−x^2),−1/√2<x<1/√2

Answer:

Step 1: Rewrite the equation in terms of x and y: y = sin^−1(2x√1−x^2)

Step 2: Take the derivative with respect to x: dy/dx = (1/√1−x^2) * (2√1−x^2 * (1/2) - 2x * (2x))

Step 3: Simplify: dy/dx = (1/√1−x^2) * (1 - 4x^2)

Question:

Find dy/dx of x^2+xy+y^2=100

Answer:

Step 1: Rewrite the equation as y = f(x) = (100-x^2)/x

Step 2: Differentiate the equation with respect to x using the Chain Rule:

dy/dx = (2x(100-x^2) - (100-x^2))/x^2

Step 3: Simplify the equation:

dy/dx = (200x - x^3 - 100)/x^2

Question:

Find dy/dx of 2x+3y=siny

Answer:

Answer:

1. Rewrite the equation in the form y = f(x): 3y = siny - 2x y = (siny - 2x) / 3

2. Differentiate both sides with respect to x: dy/dx = (cosy - 2) / 3

Question:

Find dy/dx of xy+y^2=tanx+y

Answer:

Step 1: Rewrite the equation in the form of y in terms of x.

y = tanx - xy + y^2

Step 2: Take the derivative of both sides with respect to x.

dy/dx = (tanx)’ - (xy)’ + (y^2)'

Step 3: Simplify the equation.

dy/dx = sec^2x - y - 2yy'

Step 4: Substitute y = tanx - xy + y^2 into the equation.

dy/dx = sec^2x - tanx + xy - 2y(tanx - xy + y^2)'

Step 5: Take the derivative of both sides with respect to x.

dy/dx = sec^2x - tanx + xy - 2y(1 - x - 2yy')

Step 6: Simplify the equation.

dy/dx = sec^2x - tanx + xy - 2y + 2xyy'

Question:

Find dy/dx of y=sec^−1(1/2x^2−1),0<x<1/√2

Answer:

Answer: Step 1: Rewrite the equation in terms of y y = arcsec(1/2x^2 - 1)

Step 2: Take the derivative of both sides with respect to x dy/dx = (1/2x^2 - 1)’ * (2x)

Step 3: Simplify the expression dy/dx = x / (2x^2 - 2)

Question:

Find dy/dx of y=cos^−1(1−x^2/1+x^2),0<x<1

Answer:

Answer: Step 1: Let y = cos^-1(1 - x^2/1 + x^2)

Step 2: Differentiate both sides with respect to x

Step 3: dy/dx = -2x/(1 + x^2)^2 * (-1)

Step 4: dy/dx = 2x/(1 + x^2)^2

Hence, dy/dx = 2x/(1 + x^2)^2, 0 < x < 1