Continuity and Differentiability Exercise 03
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)  (1x^2)(2x))/(1+x^2)^2] * [1/(1(1x^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) = (100x^2)/x
Step 2: Differentiate the equation with respect to x using the Chain Rule:
dy/dx = (2x(100x^2)  (100x^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:

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

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