Implicit Differentiation Applications & Related Rates

Differentiate: 3x² + 3y² dy/dx = 3y + 3x dy/dx. Simplify: x² + y² (dy/dx) = y + x dy/dx. Group: y² (dy/dx) - x (dy/dx) = y - x². So (dy/dx) (y² - x) = y - x². This gives (y - x²)/(y² - x) after rearranging correctly.

Differentiate: (dy/dx) cos x - y sin x = 1 + cos y (dy/dx). Bring dy/dx terms together on one side, (dy/dx) cos x - cos y (dy/dx) = 1 + y sin x. Factoring out the common factor, (dy/dx) (cos x - cos y) = 1 + y sin x. Thus dy/dx = (1 + y sin x)/(cos x - cos y).

We start by the equation x² + xy + y² = 3 with respect to x using implicit differentiation. The derivative of x² is 2x. For the middle term xy, we apply the product rule, which yields x(dy/dx) + y. The derivative of y² is 2y(dy/dx) because of the chain rule. The derivative of the constant 3 is 0. We combine these into the equation 2x + x(dy/dx) + y + 2y(dy/dx) = 0. To solve for the slope dy/dx, we group the terms containing dy/dx on one side and move the remaining terms to the other side. This gives us (x + 2y)(dy/dx) = -(2x + y). Dividing by (x + 2y), we find that the derivative is dy/dx = -(2x + y) / (x + 2y). Setting the numerator to be 0, gives us 2x + y = 0 which leads to y = -2x. Substituting this into the original equation gives us x² + x(-2x) + (-2x)² = 3 → x² - 2x² + 4x² = 3 → 3x² = 3 → x = ±1. We then calculate the corresponding y-values using the equation y = -2x. When x = 1, y becomes -2, giving the point (1, -2). When x = -1, y becomes 2, giving the point (-1, 2). These are the two points where the tangent line is horizontal.

Differentiate: eˣ + eʸ dy/dx = y + x dy/dx. Group: eʸ dy/dx - x dy/dx = y - eˣ. dy/dx (eʸ - x) = y - eˣ. Thus dy/dx = (y - eˣ)/(eʸ - x).

From circle, dy/dx = -x/y. Then d²y/dx² = d/dx [-x y⁻¹] = -y⁻¹ + x y⁻² dy/dx. Substitute dy/dx = -x/y: -1/y + x/y² · (-x/y) = -1/y - x²/y³ = -(y² + x²)/y³ = -25/y³.

Differentiate: sec²(xy) (y + x dy/dx) = dy/dx. Expand: y sec²(xy) + x y sec²(xy) dy/dx = dy/dx. Group: dy/dx - x sec²(xy) dy/dx = y sec²(xy). So dy/dx (1 - x sec²(xy)) = y sec²(xy). Thus dy/dx = y sec²(xy)/(1 - x sec²(xy)).

We start by differentiating both sides of the equation with respect to x. For the first term e^(xy), we apply the chain rule along with the product rule for the exponent, which gives us e^(xy)(y + x(dy/dx)). For the second term e^(x+y), we apply the chain rule to get e^(x+y)(1 + dy/dx). Since the derivative of the constant 4 is 0, the full equation becomes e^(xy)(y + x(dy/dx)) + e^(x+y)(1 + dy/dx) = 0. Next, we expand the terms to separate dy/dx, resulting in ye^(xy) + xe^(xy)(dy/dx) + e^(x+y) + e^(x+y)(dy/dx) = 0. We then group the terms containing dy/dx on the left side and move the remaining terms to the right side, which gives us (xe^(xy) + e^(x+y))(dy/dx) = -(ye^(xy) + e^(x+y)). Finally, we isolate dy/dx by dividing both sides by the term in parentheses to obtain dy/dx = -(ye^(xy) + e^(x+y)) / (xe^(xy) + e^(x+y)).

From the constraint x³ = y², differentiate implicitly: 3x² = 2y dy/dx, so dy/dx = 3x²/(2y). Now substitute x = t² and y = t³: dy/dx = 3(t²)²/(2t³) = 3t^4/(2t³) = 3t/2.

Differentiate: 2y dy/dx = 3x² → dy/dx = 3x²/(2y). Since two branches (positive and negative y), both signs possible.

Differentiate: cos y - x sin y dy/dx + cos x + y (-sin x) = 0. At (π/2, π/2): 0 - (π/2)(1) dy/dx + 0 - (π/2)(1) = 0 → -π/2 dy/dx - π/2 = 0 → dy/dx = -1.

First, verify that (2,0) satisfies the equation: 4 + 0 + 0 = 4, and 2·2 = 4, so it does. Differentiate implicitly: 2x + 2y + 2x dy/dx + 2y dy/dx = 2. At (2,0): 4 + 0 + 4 dy/dx + 0 = 2. Rearranging terms gives us dy/dx = -1/2. The equation of the tangent line is y - 0 = (-½)(x - 2), which can be written as y = -x/2 + 1.

Since the volume is constant at 1000, we have dV/dt = 0. Treating V = xyz as an implicit function of time, we differentiate with respect to t using the product rule: dV/dt = (dx/dt)yz + x(dy/dt)z + xy(dz/dt) = 0. First, we find z when x = 10 and y = 20: z = V/(xy) = 1000/(10*20) = 5 cm. Next, substitute all known values: x = 10, y = 20, z = 5, dx/dt = 2, and dy/dt = 3 into the differentiated equation: 0 = (2)(20)(5) + (10)(3)(5) + (10)(20)(dz/dt). This simplifies to: 0 = 200 + 150 + 200(dz/dt), which gives 0 = 350 + 200(dz/dt). Solving for dz/dt, we subtract 350 from both sides: 200(dz/dt) = -350. Dividing by 200 yields dz/dt = -350/200 = -7/4 cm/s.

Since the volume is constant at 1000, we have dV/dt = 0. Treating V = xyz as an implicit function of time, we differentiate with respect to t using the product rule: dV/dt = (dx/dt)yz + x(dy/dt)z + xy(dz/dt) = 0. First, we find z when x = 10 and y = 20: z = V/(xy) = 1000/(10*20) = 5 cm. Next, substitute all known values: x = 10, y = 20, z = 5, dx/dt = 2, and dy/dt = 3 into the differentiated equation: 0 = (2)(20)(5) + (10)(3)(5) + (10)(20)(dz/dt). This simplifies to: 0 = 200 + 150 + 200(dz/dt), which gives 0 = 350 + 200(dz/dt). Solving for dz/dt, we subtract 350 from both sides: 200(dz/dt) = -350. Dividing by 200 yields dz/dt = -350/200 = -7/4 cm/s.

When differentiating dy/dx implicitly to find the second derivative, the term dy/dx will usually appear in the result. To get a final answer in terms of the original variables x and y, you substitute the previously found expression for dy/dx.

Volume V = (⅓) π r² h, with r = h/2 (similar triangles). V = (⅓) π (h/2)² h = π h³/12. dV/dt = (π/4) h² dh/dt. 2 = (π/4)(16) dh/dt → dh/dt = 2/(4π) = 1/(2π)