Derivatives of Implicit Functions

Harit Thakuri

Instructor

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Basic Questions

  1. Find dydx\frac{dy}{dx} if x2+y2=25.
  2. Differentiate implicitly to find dydx\frac{dy}{dx}: xy=10xy = 10
  3. If y3+x3=6y^3 + x^3 = 6, use implicit differentiation to find dydx\frac{dy}{dx}.

Intermediate Questions

  1. Use implicit differentiation to find dydx\frac{dy}{dx} for x2+2xy+y2=12x^2 + 2xy + y^2 = 12.
  2. Differentiate implicitly: sin⁡(xy)=x2−y2\sin(xy) = x^2 - y^2.
  3. Find dydx\frac{dy}{dx} if ex⋅y2=ln⁡(x+y)e^x \cdot y^2 = \ln(x+y).

Advanced Questions

  1. For the curve defined by tan⁡(xy)=x+y\tan(xy) = x + y, find dydx\frac{dy}{dx}.
  2. Use implicit differentiation to find the slope of the tangent line to the curve x2y+ln⁡(y)=5xx^2y + \ln(y) = 5x at the point (1,e)(1, e).
  3. Find dydx\frac{dy}{dx} for the equation x3+y3−3xy=0x^3 + y^3 - 3xy = 0.
  4. If yy is implicitly defined as a function of xx by y2+xy+cos⁡(y)=4y^2 + xy + \cos(y) = 4, find dydx\frac{dy}{dx}.

Challenge Questions

  1. Given the curve x2+y2+xy=7x^2 + y^2 + xy = 7, use implicit differentiation to find dydx\frac{dy}{dx} and determine the slope at the point (2,1)(2, 1).
  2. If x2ey+y2ex=1x^2e^y + y^2e^x = 1, find dydx\frac{dy}{dx}.
  3. Differentiate implicitly to find dydx\frac{dy}{dx} for y2=x+ln⁡(y)y^2 = x + \ln(y).
  4. The curve defined by x2y2+x3+y3=10x^2y^2 + x^3 + y^3 = 10 requires implicit differentiation to find dydx\frac{dy}{dx}.
  5. Find the second derivative d2ydx2\frac{d^2y}{dx^2} for the equation x2+y2=25x^2 + y^2 = 25.

  • 30 Dec, 2024
Replies (1)
Harit Thakuri

Instructor

  • 22
  • 0

Basic Questions

1. x2+y2=25x^2 + y^2 = 25
Differentiate both sides w.r.t. xx:

2x+2ydydx=02x + 2y \frac{dy}{dx} = 0

Solve for dydx\frac{dy}{dx}:

dydx=−xy\frac{dy}{dx} = -\frac{x}{y}

2. xy=10xy = 10
Differentiate both sides w.r.t. xx:

xdydx+y=0x\frac{dy}{dx} + y = 0

Solve for dydx\frac{dy}{dx}:

dydx=−yx\frac{dy}{dx} = -\frac{y}{x}

3. y3+x3=6y^3 + x^3 = 6
Differentiate both sides w.r.t. xx:

3y2dydx+3x2=03y^2 \frac{dy}{dx} + 3x^2 = 0

Solve for dydx\frac{dy}{dx}:

dydx=−x2y2\frac{dy}{dx} = -\frac{x^2}{y^2}

Intermediate Questions

4. x2+2xy+y2=12x^2 + 2xy + y^2 = 12
Differentiate both sides w.r.t. xx:

2x+2xdydx+2y+2ydydx=02x + 2x\frac{dy}{dx} + 2y + 2y\frac{dy}{dx} = 0

Simplify:

2x+2y+(2x+2y)dydx=02x + 2y + (2x + 2y)\frac{dy}{dx} = 0

Solve for dydx\frac{dy}{dx}:

dydx=−x+yx+y\frac{dy}{dx} = -\frac{x + y}{x + y}5.

sin⁡(xy)=x2−y2\sin(xy) = x^2 - y^2
Differentiate both sides w.r.t. xx:

cos⁡(xy)⋅(y+xdydx)=2x−2ydydx\cos(xy) \cdot \left(y + x \frac{dy}{dx}\right) = 2x - 2y \frac{dy}{dx}

Expand:

cos⁡(xy)⋅y+cos⁡(xy)⋅xdydx=2x−2ydydx\cos(xy) \cdot y + \cos(xy) \cdot x \frac{dy}{dx} = 2x - 2y \frac{dy}{dx}

Group terms involving dydx\frac{dy}{dx}:

cos⁡(xy)⋅xdydx+2ydydx=2x−cos⁡(xy)⋅y\cos(xy) \cdot x \frac{dy}{dx} + 2y \frac{dy}{dx} = 2x - \cos(xy) \cdot y

Factor out dydx\frac{dy}{dx}:

dydx(cos⁡(xy)⋅x+2y)=2x−cos⁡(xy)⋅y\frac{dy}{dx} \left(\cos(xy) \cdot x + 2y\right) = 2x - \cos(xy) \cdot y

Solve for dydx\frac{dy}{dx}:

dydx=2x−cos⁡(xy)⋅ycos⁡(xy)⋅x+2y\frac{dy}{dx} = \frac{2x - \cos(xy) \cdot y}{\cos(xy) \cdot x + 2y}

6. ex⋅y2=ln⁡(x+y)e^x \cdot y^2 = \ln(x + y)
Differentiate both sides w.r.t. xx:

ex⋅y2+ex⋅2ydydx=1x+y⋅(1+dydx)e^x \cdot y^2 + e^x \cdot 2y \frac{dy}{dx} = \frac{1}{x + y} \cdot \left(1 + \frac{dy}{dx}\right)

Expand:

ex⋅y2+2ex⋅ydydx=1x+y+1x+ydydxe^x \cdot y^2 + 2e^x \cdot y \frac{dy}{dx} = \frac{1}{x + y} + \frac{1}{x + y} \frac{dy}{dx}

Group terms involving dydx\frac{dy}{dx}:

2ex⋅ydydx−1x+ydydx=1x+y−ex⋅y22e^x \cdot y \frac{dy}{dx} - \frac{1}{x + y} \frac{dy}{dx} = \frac{1}{x + y} - e^x \cdot y^2

Factor out dydx\frac{dy}{dx}:

dydx(2ex⋅y−1x+y)=1x+y−ex⋅y2\frac{dy}{dx} \left(2e^x \cdot y - \frac{1}{x + y}\right) = \frac{1}{x + y} - e^x \cdot y^2

Solve for dydx\frac{dy}{dx}:

dydx=1x+y−ex⋅y22ex⋅y−1x+y\frac{dy}{dx} = \frac{\frac{1}{x + y} - e^x \cdot y^2}{2e^x \cdot y - \frac{1}{x + y}}

Advanced Questions

7. tan⁡(xy)=x+y\tan(xy) = x + y
Differentiate both sides w.r.t. xx:

sec⁡2(xy)⋅(y+xdydx)=1+dydx\sec^2(xy) \cdot \left(y + x \frac{dy}{dx}\right) = 1 + \frac{dy}{dx}

Expand:

sec⁡2(xy)⋅y+sec⁡2(xy)⋅xdydx=1+dydx\sec^2(xy) \cdot y + \sec^2(xy) \cdot x \frac{dy}{dx} = 1 + \frac{dy}{dx}

Group terms involving dydx\frac{dy}{dx}:

sec⁡2(xy)⋅xdydx−dydx=1−sec⁡2(xy)⋅y\sec^2(xy) \cdot x \frac{dy}{dx} - \frac{dy}{dx} = 1 - \sec^2(xy) \cdot y

Factor out dydx\frac{dy}{dx}:

dydx(sec⁡2(xy)⋅x−1)=1−sec⁡2(xy)⋅y\frac{dy}{dx} \left(\sec^2(xy) \cdot x - 1\right) = 1 - \sec^2(xy) \cdot y

Solve for dydx\frac{dy}{dx}:

dydx=1−sec⁡2(xy)⋅ysec⁡2(xy)⋅x−1\frac{dy}{dx} = \frac{1 - \sec^2(xy) \cdot y}{\sec^2(xy) \cdot x - 1}

8. x2y+ln⁡(y)=5xx^2y + \ln(y) = 5x
Differentiate both sides w.r.t. xx:

2xy+x2dydx+1ydydx=52xy + x^2 \frac{dy}{dx} + \frac{1}{y} \frac{dy}{dx} = 5

Group terms involving dydx\frac{dy}{dx}:

x2dydx+1ydydx=5−2xyx^2 \frac{dy}{dx} + \frac{1}{y} \frac{dy}{dx} = 5 - 2xy

Factor out dydx\frac{dy}{dx}:

dydx(x2+1y)=5−2xy\frac{dy}{dx} \left(x^2 + \frac{1}{y}\right) = 5 - 2xy

Solve for dydx\frac{dy}{dx}:

dydx=5−2xyx2+1y\frac{dy}{dx} = \frac{5 - 2xy}{x^2 + \frac{1}{y}}

9. x3+y3−3xy=0x^3 + y^3 - 3xy = 0
Differentiate both sides w.r.t. xx

3x2+3y2dydx−3(y+xdydx)=03x^2 + 3y^2 \frac{dy}{dx} - 3\left(y + x \frac{dy}{dx}\right) = 0

Expand:

3x2+3y2dydx−3y−3xdydx=03x^2 + 3y^2 \frac{dy}{dx} - 3y - 3x \frac{dy}{dx} = 0

Group terms involving dydx\frac{dy}{dx}:

3y2dydx−3xdydx=3y−3x23y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} = 3y - 3x^2

Factor out dydx\frac{dy}{dx}:

dydx(3y2−3x)=3y−3x2\frac{dy}{dx} \left(3y^2 - 3x\right) = 3y - 3x^2

Solve for dydx\frac{dy}{dx}:

dydx=y−x2y2−x\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}

10. y2+xy+cos⁡(y)=4y^2 + xy + \cos(y) = 4
Differentiate both sides w.r.t. xx:

2ydydx+y+xdydx−sin⁡(y)dydx=02y \frac{dy}{dx} + y + x \frac{dy}{dx} - \sin(y) \frac{dy}{dx} = 0

Group terms involving dydx\frac{dy}{dx}:

dydx(2y+x−sin⁡(y))=−y\frac{dy}{dx} \left(2y + x - \sin(y)\right) = -y

Solve for dydx\frac{dy}{dx}:

dydx=−y2y+x−sin⁡(y)\frac{dy}{dx} = \frac{-y}{2y + x - \sin(y)}

Challenge Questions

11. x2+y2+xy=7x^2 + y^2 + xy = 7
Differentiate both sides w.r.t. xx:

2x+2ydydx+y+xdydx=02x + 2y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0

Group terms involving dydx\frac{dy}{dx}:

(2y+x)dydx=−2x−y(2y + x) \frac{dy}{dx} = -2x - y

Solve for dydx\frac{dy}{dx}:

dydx=−2x−y2y+x\frac{dy}{dx} = \frac{-2x - y}{2y + x}

At (2,1)(2, 1):

dydx=−2(2)−12(1)+2=−4−12+2=−54\frac{dy}{dx} = \frac{-2(2) - 1}{2(1) + 2} = \frac{-4 - 1}{2 + 2} = \frac{-5}{4}

12. x2ey+y2ex=1x^2 e^y + y^2 e^x = 1
Differentiate both sides w.r.t. xx:

2xey+x2eydydx+2yexdydx+y2ex=02x e^y + x^2 e^y \frac{dy}{dx} + 2y e^x \frac{dy}{dx} + y^2 e^x = 0

Group terms involving dydx\frac{dy}{dx}:

x2eydydx+2yexdydx=−2xey−y2exx^2 e^y \frac{dy}{dx} + 2y e^x \frac{dy}{dx} = -2x e^y - y^2 e^x

Factor out dydx\frac{dy}{dx}:

dydx(x2ey+2yex)=−2xey−y2ex\frac{dy}{dx} \left(x^2 e^y + 2y e^x\right) = -2x e^y - y^2 e^x

Solve for dydx\frac{dy}{dx}:

dydx=−2xey−y2exx2ey+2yex\frac{dy}{dx} = \frac{-2x e^y - y^2 e^x}{x^2 e^y + 2y e^x}

  • 30 Dec, 2024

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