If $z=f(x+ay)+\phi(x-ay)$ then :

Question

Accepted Answer

To determine why Option 2 is the correct answer, we need to analyze the given function $ z = f(x + ay) + \phi(x - ay) $ and compute the second partial derivatives with respect to $ x $ and $ y $.

### Explanation for Option 2:

1. **Function Analysis**:
   - Let $ u = x + ay $ and $ v = x - ay $.
   - Then, $ z = f(u) + \phi(v) $.

2. **First Partial Derivatives**:
   - $\frac{\partial z}{\partial x} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial \phi}{\partial v} \cdot \frac{\partial v}{\partial x} = f'(u) + \phi'(v)$.
   - $\frac{\partial z}{\partial y} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial y} + \frac{\partial \phi}{\partial v} \cdot \frac{\partial v}{\partial y} = af'(u) - a\phi'(v)$.

3. **Second Partial Derivatives**:
   - $\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x}(f'(u) + \phi'(v)) = f''(u) + \phi''(v)$.
   - $\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y}(af'(u) - a\phi'(v)) = a^2f''(u) + a^2\phi''(v)$.

4. **Relation Between Second Partial Derivatives**:
   - From the above, $\frac{\partial^2 z}{\partial x^2} = f''(u) + \phi''(v)$.
   - $\frac{\partial^2 z}{\partial y^2} = a^2(f''(u) + \phi''(v))$.

Therefore, $\frac{\partial^2 z}{\partial y^2} = a^2 \frac{\partial^2 z}{\partial x^2}$.

This matches Option 2: $\frac{\partial^2 z}{\partial y^2} = a^2 \frac{\partial^2 z}{\partial x^2}$.

### Explanation for Why Other Options are Incorrect:

- **Option 1**: $\frac{\partial^2 z}{\partial x^2} = a^2 \frac{\partial^2 z}{\partial y^2}$ is incorrect because it suggests the inverse relationship of what we derived. The correct relationship is $\frac{\partial^2 z}{\partial y^2} = a^2 \frac{\partial^2 z}{\partial x^2}$.

- **Option 3**: $\frac{\partial^2 z}{\partial x^2} = \frac{-1}{a^2} \frac{\partial^2 z}{\partial y^2}$ is incorrect because it introduces a negative sign and an inverse square factor that does not appear in the correct relationship.

- **Option 4**: $\frac{\partial^2 z}{\partial x^2} = 2a^2 \frac{\partial^2 z}{\partial y^2}$ is incorrect because it incorrectly introduces a factor of 2, which is not present in the derived relationship.

In conclusion, Option 2 is the correct answer because it accurately reflects the relationship between the second partial derivatives of $ z $ with respect to $ x $ and $ y $ as derived from the given function.