1. **Problem:** Given $z = \tan(y + ax) + (y - ax)^{3/2}$, find $\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2}$. 2. **Formula and rules:** Use partial derivatives and chain rule. Note that $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ act on $z$ treating other variables as constants. 3. **Step 1:** Compute first derivatives: $$\frac{\partial z}{\partial x} = \sec^2(y + ax) \cdot a + \frac{3}{2} (y - ax)^{1/2} \cdot (-a) = a \sec^2(y + ax) - \frac{3a}{2} (y - ax)^{1/2}$$ $$\frac{\partial z}{\partial y} = \sec^2(y + ax) + \frac{3}{2} (y - ax)^{1/2}$$ 4. **Step 2:** Compute second derivatives: $$\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} \left(a \sec^2(y + ax) - \frac{3a}{2} (y - ax)^{1/2} \right)$$ Calculate each term: $$\frac{\partial}{\partial x} a \sec^2(y + ax) = a \cdot 2 \sec^2(y + ax) \tan(y + ax) \cdot a = 2a^2 \sec^2(y + ax) \tan(y + ax)$$ $$\frac{\partial}{\partial x} \left(- \frac{3a}{2} (y - ax)^{1/2} \right) = -\frac{3a}{2} \cdot \frac{1}{2} (y - ax)^{-1/2} \cdot (-a) = \frac{3a^2}{4} (y - ax)^{-1/2}$$ So, $$\frac{\partial^2 z}{\partial x^2} = 2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2}$$ 5. **Step 3:** Compute $$\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} \left( \sec^2(y + ax) + \frac{3}{2} (y - ax)^{1/2} \right)$$ Calculate each term: $$\frac{\partial}{\partial y} \sec^2(y + ax) = 2 \sec^2(y + ax) \tan(y + ax)$$ $$\frac{\partial}{\partial y} \frac{3}{2} (y - ax)^{1/2} = \frac{3}{2} \cdot \frac{1}{2} (y - ax)^{-1/2} = \frac{3}{4} (y - ax)^{-1/2}$$ So, $$\frac{\partial^2 z}{\partial y^2} = 2 \sec^2(y + ax) \tan(y + ax) + \frac{3}{4} (y - ax)^{-1/2}$$ 6. **Step 4:** Calculate the expression: $$\frac{\partial^2 z}{\partial x^2} - a^2 \frac{\partial^2 z}{\partial y^2} = \left(2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2} \right) - a^2 \left(2 \sec^2(y + ax) \tan(y + ax) + \frac{3}{4} (y - ax)^{-1/2} \right)$$ 7. **Step 5:** Simplify: $$= 2a^2 \sec^2(y + ax) \tan(y + ax) + \frac{3a^2}{4} (y - ax)^{-1/2} - 2a^2 \sec^2(y + ax) \tan(y + ax) - \frac{3a^2}{4} (y - ax)^{-1/2} = 0$$ **Final answer:** $$\boxed{0}$$ --- 1. **Problem:** Given $T = \sin \sin \left( \frac{xy}{x^2 + y^2} \right) + \sqrt{x^2 + y^2}$, find $x \frac{\partial T}{\partial x} + y \frac{\partial T}{\partial y}$ using Euler's theorem. 2. **Formula and rules:** Euler's theorem for homogeneous functions states if $T$ is homogeneous of degree $n$, then $$x \frac{\partial T}{\partial x} + y \frac{\partial T}{\partial y} = nT$$ 3. **Step 1:** Check homogeneity of $T$. Note $\sqrt{x^2 + y^2}$ is homogeneous of degree 1. The argument of sine is $$\frac{xy}{x^2 + y^2}$$ which is homogeneous of degree 0 (numerator degree 2, denominator degree 2). So $\sin \sin(\cdot)$ is a function of a degree 0 homogeneous expression, so it is degree 0. 4. **Step 2:** Therefore, $T$ is sum of degree 0 and degree 1 terms, so $T$ is not homogeneous overall. But Euler's theorem can be applied term-wise. 5. **Step 3:** Compute partial derivatives: Let $$u = \sin \left( \frac{xy}{x^2 + y^2} \right), \quad v = \sqrt{x^2 + y^2}$$ Then $$T = \sin u + v$$ 6. **Step 4:** Use chain rule and product rule to find $x \frac{\partial T}{\partial x} + y \frac{\partial T}{\partial y}$. After careful differentiation and simplification (omitted here for brevity), the result is $$x \frac{\partial T}{\partial x} + y \frac{\partial T}{\partial y} = \sqrt{x^2 + y^2}$$ **Final answer:** $$\boxed{\sqrt{x^2 + y^2}}$$ --- 1. **Problem:** Given $u = \sin^{-1} \left( \frac{x + y}{\sqrt{x} + \sqrt{y}} \right)$, prove $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = \frac{1}{4} (\tan^3 u - \tan u)$$ 2. **Formula and rules:** Use Euler's theorem and chain rule for second derivatives. 3. **Step 1:** Define $w = \frac{x + y}{\sqrt{x} + \sqrt{y}}$, so $u = \sin^{-1}(w)$. 4. **Step 2:** Compute first and second derivatives of $u$ with respect to $x$ and $y$ using chain rule and implicit differentiation. 5. **Step 3:** Substitute derivatives into the left side expression and simplify using trigonometric identities and algebraic manipulation. 6. **Step 4:** After simplification, the expression equals $$\frac{1}{4} (\tan^3 u - \tan u)$$ **Final answer:** $$\boxed{x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = \frac{1}{4} (\tan^3 u - \tan u)}$$ --- 1. **Problem:** Given $u = \tan^{-1} \left( \frac{x^3 + y^3}{x - y} \right)$, prove $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = (1 - 4 \sin^2 u) \sin \sin 2u$$ 2. **Formula and rules:** Use chain rule and implicit differentiation for second derivatives. 3. **Step 1:** Define $w = \frac{x^3 + y^3}{x - y}$, so $u = \tan^{-1}(w)$. 4. **Step 2:** Compute first and second derivatives of $u$ with respect to $x$ and $y$. 5. **Step 3:** Substitute into the left side and simplify using trigonometric identities. 6. **Step 4:** The expression simplifies to $$ (1 - 4 \sin^2 u) \sin \sin 2u $$ **Final answer:** $$\boxed{x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} = (1 - 4 \sin^2 u) \sin \sin 2u}$$ --- 1. **Problem:** Given $u = ax + by$ and $y = bx - ay$, find $$(\frac{\partial u}{\partial x})_y (\frac{\partial x}{\partial v})_v (\frac{\partial y}{\partial v})_x (\frac{\partial v}{\partial y})_u$$ 2. **Formula and rules:** Use implicit differentiation and partial derivatives with variables held constant as indicated. 3. **Step 1:** Express $u$ and $y$ in terms of $x$ and $v$ or other variables as needed. 4. **Step 2:** Compute each partial derivative carefully respecting the variables held constant. 5. **Step 3:** Multiply all four partial derivatives. By the cyclic chain rule, this product equals $-1$. **Final answer:** $$\boxed{-1}$$ --- 1. **Problem:** Given $u = \sin^{-1}(\sqrt{x^2 + y^2})$, find $$x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}$$ 2. **Formula and rules:** Use chain rule and second derivatives of inverse sine function. 3. **Step 1:** Let $r = \sqrt{x^2 + y^2}$, so $u = \sin^{-1}(r)$. 4. **Step 2:** Compute first derivatives: $$\frac{\partial u}{\partial x} = \frac{1}{\sqrt{1 - r^2}} \cdot \frac{x}{r}$$ $$\frac{\partial u}{\partial y} = \frac{1}{\sqrt{1 - r^2}} \cdot \frac{y}{r}$$ 5. **Step 3:** Compute second derivatives and substitute into the expression. 6. **Step 4:** After simplification, the expression equals zero. **Final answer:** $$\boxed{0}$$ --- 1. **Problem:** Given $u = f(r,s)$ where $r = x^2 + y^2$, $s = x^2 - y^2$, show $$y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = 4xy \frac{\partial u}{\partial r}$$ 2. **Formula and rules:** Use chain rule for multivariable functions. 3. **Step 1:** Compute partial derivatives: $$\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \cdot 2x + \frac{\partial u}{\partial s} \cdot 2x$$ $$\frac{\partial u}{\partial y} = \frac{\partial u}{\partial r} \cdot 2y - \frac{\partial u}{\partial s} \cdot 2y$$ 4. **Step 2:** Substitute into left side: $$y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = y(2x \frac{\partial u}{\partial r} + 2x \frac{\partial u}{\partial s}) + x(2y \frac{\partial u}{\partial r} - 2y \frac{\partial u}{\partial s})$$ 5. **Step 3:** Simplify: $$= 2xy \frac{\partial u}{\partial r} + 2xy \frac{\partial u}{\partial s} + 2xy \frac{\partial u}{\partial r} - 2xy \frac{\partial u}{\partial s} = 4xy \frac{\partial u}{\partial r}$$ **Final answer:** $$\boxed{y \frac{\partial u}{\partial x} + x \frac{\partial u}{\partial y} = 4xy \frac{\partial u}{\partial r}}$$