24. $$ \cos (x y)+\sin y=x y \quad $$ 25. $$ 2 x e^{x}+2 y e^{y}=4 $$ 26. If $$ f(x)+x^{2}[f(x)]^{3}=10 $$ and $$ f(1)=2 $$, find $$ f^{\prime}(1) $$. 27. If $$ g(x)+x \sin g(x)=x^{2} $$, find $$ g^{\prime}(0) $$. Assume $$ y $$ is the independent variable and $$ x $$ is the dependent variable and use implicit differentiation to find $$ \frac{d x}{d y} $$. 28. $$ x^{4} y^{2}-x^{3} y+2 x y^{3}=0 $$ 29. $$ y \sec x=x an y $$ Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 30. $$ x^{2} y+2 x=15, \quad(3,1) $$ 31. $$ y \sin 2 x=x \cos 2 y, \quad\left(\frac{\pi}{2}, \frac{\pi}{4} ight) $$ 32. $$ \sin (x+y)=2 x-2 y, \quad(\pi, \pi) $$ 33. $$ x^{2}-x y-y^{2}=1, \quad(2,1) $$ 34. $$ x^{2}+2 x y+4 y^{2}=12, \quad(2,1) $$

Question

24. $$ \cos (x y)+\sin y=x y \quad $$ 25. $$ 2 x e^{x}+2 y e^{y}=4 $$ 26. If $$ f(x)+x^{2}[f(x)]^{3}=10 $$ and $$ f(1)=2 $$, find $$ f^{\prime}(1) $$. 27. If $$ g(x)+x \sin g(x)=x^{2} $$, find $$ g^{\prime}(0) $$. Assume $$ y $$ is the independent variable and $$ x $$ is the dependent variable and use implicit differentiation to find $$ \frac{d x}{d y} $$. 28. $$ x^{4} y^{2}-x^{3} y+2 x y^{3}=0 $$ 29. $$ y \sec x=x 	an y $$ Use implicit differentiation to find an equation of the tangent line to the curve at the given point. 30. $$ x^{2} y+2 x=15, \quad(3,1) $$ 31. $$ y \sin 2 x=x \cos 2 y, \quad\left(\frac{\pi}{2}, \frac{\pi}{4}ight) $$ 32. $$ \sin (x+y)=2 x-2 y, \quad(\pi, \pi) $$ 33. $$ x^{2}-x y-y^{2}=1, \quad(2,1) $$ 34. $$ x^{2}+2 x y+4 y^{2}=12, \quad(2,1) $$

UpStudy AI · Accepted Answer

Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$x^{2}y+2x=15$$
- step1: Move the expression to the right side:
  $$x^{2}y=15-2x$$
- step2: Divide both sides:
  $$\frac{x^{2}y}{x^{2}}=\frac{15-2x}{x^{2}}$$
- step3: Divide the numbers:
  $$y=\frac{15-2x}{x^{2}}$$
Solve the equation $$ x^{2}-xy-y^{2}=1 $$.
Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$x^{2}-xy-y^{2}=1$$
- step1: Move the expression to the left side:
  $$x^{2}-xy-y^{2}-1=0$$
- step2: Simplify:
  $$x^{2}-1-xy-y^{2}=0$$
- step3: Rewrite in standard form:
  $$-y^{2}-xy+x^{2}-1=0$$
- step4: Multiply both sides:
  $$y^{2}+xy-x^{2}+1=0$$
- step5: Solve using the quadratic formula:
  $$y=\frac{-x\pm \sqrt{x^{2}-4\left(-x^{2}+1\right)}}{2}$$
- step6: Simplify the expression:
  $$y=\frac{-x\pm \sqrt{5x^{2}-4}}{2}$$
- step7: Separate into possible cases:
  $$\begin{align}&y=\frac{-x+\sqrt{5x^{2}-4}}{2}\&y=\frac{-x-\sqrt{5x^{2}-4}}{2}\end{align}$$
- step8: Rewrite the fraction:
  $$\begin{align}&y=\frac{-x+\sqrt{5x^{2}-4}}{2}\&y=-\frac{x+\sqrt{5x^{2}-4}}{2}\end{align}$$
Solve the equation $$ x^{2}+2xy+4y^{2}=12 $$.
Solve the equation by following steps:
- step0: Solve for $$y$$:
  $$x^{2}+2xy+4y^{2}=12$$
- step1: Move the expression to the left side:
  $$x^{2}+2xy+4y^{2}-12=0$$
- step2: Simplify:
  $$x^{2}-12+2xy+4y^{2}=0$$
- step3: Rewrite in standard form:
  $$4y^{2}+2xy+x^{2}-12=0$$
- step4: Solve using the quadratic formula:
  $$y=\frac{-2x\pm \sqrt{\left(2x\right)^{2}-4\times 4\left(x^{2}-12\right)}}{2\times 4}$$
- step5: Simplify the expression:
  $$y=\frac{-2x\pm \sqrt{\left(2x\right)^{2}-4\times 4\left(x^{2}-12\right)}}{8}$$
- step6: Simplify the expression:
  $$y=\frac{-2x\pm \sqrt{-12x^{2}+192}}{8}$$
- step7: Simplify the expression:
  $$y=\frac{-2x\pm 2\sqrt{48-3x^{2}}}{8}$$
- step8: Separate into possible cases:
  $$\begin{align}&y=\frac{-2x+2\sqrt{48-3x^{2}}}{8}\&y=\frac{-2x-2\sqrt{48-3x^{2}}}{8}\end{align}$$
- step9: Simplify the expression:
  $$\begin{align}&y=\frac{-x+\sqrt{48-3x^{2}}}{4}\&y=\frac{-2x-2\sqrt{48-3x^{2}}}{8}\end{align}$$
- step10: Simplify the expression:
  $$\begin{align}&y=\frac{-x+\sqrt{48-3x^{2}}}{4}\&y=-\frac{x+\sqrt{48-3x^{2}}}{4}\end{align}$$
Vamos a resolver los problemas uno por uno, extrayendo las condiciones conocidas y utilizando la diferenciación implícita donde sea necesario.

### Problema 24
**Ecuación:** $$ \cos (xy) + \sin y = xy $$

No se encontró una solución directa para esta ecuación.

### Problema 25
**Ecuación:** $$ 2xe^{x} + 2ye^{y} = 4 $$

No se encontró una solución directa para esta ecuación.

### Problema 26
**Ecuación:** $$ f(x) + x^{2}[f(x)]^{3} = 10 $$ y $$ f(1) = 2 $$

Para encontrar $$ f'(1) $$, aplicamos la diferenciación implícita:

1. Derivamos ambos lados respecto a $$ x $$:
   $$
   f'(x) + 2x f(x)^{3} + 3x^{2} f(x)^{2} f'(x) = 0
   $$

2. Sustituyendo $$ x = 1 $$ y $$ f(1) = 2 $$:
   $$
   f'(1) + 2(1)(2)^{3} + 3(1)^{2}(2)^{2} f'(1) = 0
   $$
   $$
   f'(1) + 16 + 12 f'(1) = 0
   $$
   $$
   13 f'(1) = -16 \implies f'(1) = -\frac{16}{13}
   $$

### Problema 27
**Ecuación:** $$ g(x) + x \sin g(x) = x^{2} $$

Para encontrar $$ g'(0) $$:

1. Derivamos ambos lados respecto a $$ x $$:
   $$
   g'(x) + \sin g(x) + x \cos g(x) g'(x) = 2x
   $$

2. Evaluamos en $$ x = 0 $$:
   $$
   g'(0) + \sin g(0) = 0
   $$
   Asumiendo $$ g(0) = 0 $$ (ya que $$ g(0) + 0 = 0 $$), tenemos:
   $$
   g'(0) + 0 = 0 \implies g'(0) = 0
   $$

### Problema 28
**Ecuación:** $$ x^{4}y^{2} - x^{3}y + 2xy^{3} = 0 $$

No se encontró una solución directa para esta ecuación.

### Problema 29
**Ecuación:** $$ y \sec x = x \tan y $$

Para encontrar la derivada implícita:

1. Derivamos ambos lados respecto a $$ x $$:
   $$
   \sec x \tan x \frac{dy}{dx} + y \sec x \tan x = \tan y + x \sec^{2} y \frac{dy}{dx}
   $$

2. Reorganizamos para encontrar $$ \frac{dy}{dx} $$.

No se encontró una solución directa para esta ecuación.

### Problema 30
**Ecuación:** $$ x^{2}y + 2x = 15 $$ en el punto $$ (3,1) $$

1. Derivamos:
   $$
   2xy + x^{2} \frac{dy}{dx} + 2 = 0
   $$
2. Sustituyendo $$ (3,1) $$:
   $$
   2(3)(1) + 3^{2} \frac{dy}{dx} + 2 = 0 \implies 6 + 9 \frac{dy}{dx} + 2 = 0
   $$
   $$
   9 \frac{dy}{dx} = -8 \implies \frac{dy}{dx} = -\frac{8}{9}
   $$

### Problema 31
**Ecuación:** $$ y \sin 2x = x \cos 2y $$ en el punto $$ \left(\frac{\pi}{2}, \frac{\pi}{4}\right) $$

1. Derivamos:
   $$
   \sin 2x \frac{dy}{dx} + 2y \cos 2x = \cos 2y \frac{dy}{dx} - 2x \sin 2y \frac{dy}{dx}
   $$
2. Sustituyendo $$ \left(\frac{\pi}{2}, \frac{\pi}{4}\right) $$ y resolviendo para $$ \frac{dy}{dx} $$.

No se encontró una solución directa para esta ecuación.

### Problema 32
**Ecuación:** $$ \sin (x+y) = 2x - 2y $$ en el punto $$ (\pi, \pi) $$

1. Derivamos:
   $$
   \cos(x+y) \left(1 + \frac{dy}{dx}\right) = 2 - 2 \frac{dy}{dx}
   $$
2. Sustituyendo $$ (\pi, \pi) $$ y resolviendo para $$ \frac{dy}{dx} $$.

No se encontró una solución directa para esta ecuación.

### Problema 33
**Ecuación:** $$ x^{2} - xy - y^{2} = 1 $$ en el punto $$ (2,1) $$

1. Derivamos:
   $$
   2x - y - x \frac{dy}{dx} - 2y \frac{dy}{dx} = 0
   $$
2. Sustituyendo $$ (2,1) $$:
   $$
   4 - 1 - 2 \frac{dy}{dx} - 2 \frac{dy}{dx} = 0 \implies 3 = 4 \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{3}{4}
   $$

### Problema 34
**Ecuación:** $$ x^{2} + 2xy + 4y^{2} = 12 $$ en el punto $$ (2,1) $$

1. Derivamos:
   $$
   2x + 2y + 2x \frac{dy}{dx} + 8y \frac{dy}{dx} = 0
   $$
2. Sustituyendo $$ (2,1) $$:
   $$
   4 + 2 + 2(2) \frac{dy}{dx} + 8(1) \frac{dy}{dx} = 0 \implies 6 + 20 \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{3}{10}
   $$

### Problema 26 La derivada de $$ f(x) $$ en $$ x = 1 $$ es $$ f'(1) = -\frac{16}{13} $$. ### Problema 27 La derivada de $$ g(x) $$ en $$ x = 0 $$ es $$ g'(0) = 0 $$. ### Problema 30 La pendiente de la recta tangente en $$ (3,1) $$ es $$ \frac{dy}{dx} = -\frac{8}{9} $$. ### Problema 33 La pendiente de la recta tangente en $$ (2,1) $$ es $$ \frac{dy}{dx} = \frac{3}{4} $$. ### Problema 34 La pendiente de la recta tangente en $$ (2,1) $$ es $$ \frac{dy}{dx} = -\frac{3}{10} $$.

25.

If and , find .

If , find .
Assume is the independent variable and is the dependent
variable and use implicit differentiation to find .

Use implicit differentiation to find an equation of the tangent line
to the curve at the given point.

Upstudy AI Solution

Answer

Problema 26

Problema 27

Problema 30

Problema 33

Problema 34

Solution

The Deep Dive

Related Questions

Latest Calculus Questions

25. If and , find . If , find . Assume is the independent variable and is the dependent variable and use implicit differentiation to find . Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Upstudy AI Solution

Answer

Problema 26

Problema 27

Problema 30

Problema 33

Problema 34

Solution

The Deep Dive

Related Questions

Latest Calculus Questions

25.

If and , find .

If , find .
Assume is the independent variable and is the dependent
variable and use implicit differentiation to find .

Use implicit differentiation to find an equation of the tangent line
to the curve at the given point.