If \(\begin{array}{l} H = {\tan ^{ - 1}}\frac{x}{y}\\ \end{array}\), x = u + v, y = u - v then \(\begin{array}{l} \frac{{\partial H}}{{\partial V}}\\ \end{array}\)is
Calculate Partial Derivative using the Chain Rule
We are given a function \(H\) defined as \(H = {\tan ^{ - 1}}\frac{x}{y}\), where \(x\) and \(y\) are themselves functions of \(u\) and \(v\), specifically \(x = u + v\) and \(y = u - v\). We need to find the partial derivative of \(H\) with respect to \(v\), denoted as \(\frac{{\partial H}}{{\partial v}}\).
Since \(H\) is a function of \(x\) and \(y\), and \(x\) and \(y\) are functions of \(u\) and \(v\), we must use the chain rule for multivariable functions. The formula for \(\frac{{\partial H}}{{\partial v}}\) is:
\(\frac{{\partial H}}{{\partial v}} = \frac{{\partial H}}{{\partial x}} \frac{{\partial x}}{{\partial v}} + \frac{{\partial H}}{{\partial y}} \frac{{\partial y}}{{\partial v}}\)
Let's calculate each of the partial derivatives on the right side of the equation.
Step 1: Find \(\frac{{\partial H}}{{\partial x}}\)
Given \(H = \tan^{-1}\left(\frac{x}{y}\right)\). To find \(\frac{{\partial H}}{{\partial x}}\), we treat \(y\) as a constant.
\(\frac{{\partial H}}{{\partial x}} = \frac{\partial}{\partial x} \left( \tan^{-1}\left(\frac{x}{y}\right) \right)\)
Using the chain rule for single variable derivatives, \(\frac{d}{dz}(\tan^{-1}(z)) = \frac{1}{1+z^2}\), with \(z = \frac{x}{y}\), and the derivative of the inner function \(\frac{\partial}{\partial x}(\frac{x}{y}) = \frac{1}{y}\) (since \(y\) is constant with respect to \(x\)):
\(\frac{{\partial H}}{{\partial x}} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \frac{1}{y}\)
\(\frac{{\partial H}}{{\partial x}} = \frac{1}{\frac{y^2 + x^2}{y^2}} \cdot \frac{1}{y}\)
\(\frac{{\partial H}}{{\partial x}} = \frac{y^2}{x^2 + y^2} \cdot \frac{1}{y}\)
\(\frac{{\partial H}}{{\partial x}} = \frac{y}{x^2 + y^2}\)
Step 2: Find \(\frac{{\partial H}}{{\partial y}}\)
Given \(H = \tan^{-1}\left(\frac{x}{y}\right)\). To find \(\frac{{\partial H}}{{\partial y}}\), we treat \(x\) as a constant.
\(\frac{{\partial H}}{{\partial y}} = \frac{\partial}{\partial y} \left( \tan^{-1}\left(\frac{x}{y}\right) \right)\)
Using the chain rule, with \(z = \frac{x}{y}\), and the derivative of the inner function \(\frac{\partial}{\partial y}(\frac{x}{y}) = x \cdot \frac{\partial}{\partial y}(y^{-1}) = x \cdot (-1)y^{-2} = -\frac{x}{y^2}\):
\(\frac{{\partial H}}{{\partial y}} = \frac{1}{1 + \left(\frac{x}{y}\right)^2} \cdot \left(-\frac{x}{y^2}\right)\)
\(\frac{{\partial H}}{{\partial y}} = \frac{1}{\frac{y^2 + x^2}{y^2}} \cdot \left(-\frac{x}{y^2}\right)\)
\(\frac{{\partial H}}{{\partial y}} = \frac{y^2}{x^2 + y^2} \cdot \left(-\frac{x}{y^2}\right)\)
\(\frac{{\partial H}}{{\partial y}} = -\frac{x}{x^2 + y^2}\)
Step 3: Find \(\frac{{\partial x}}{{\partial v}}\)
Given \(x = u + v\). To find \(\frac{{\partial x}}{{\partial v}}\), we treat \(u\) as a constant.
\(\frac{{\partial x}}{{\partial v}} = \frac{\partial}{\partial v} (u + v) = 0 + 1 = 1\)
Step 4: Find \(\frac{{\partial y}}{{\partial v}}\)
Given \(y = u - v\). To find \(\frac{{\partial y}}{{\partial v}}\), we treat \(u\) as a constant.
\(\frac{{\partial y}}{{\partial v}} = \frac{\partial}{\partial v} (u - v) = 0 - 1 = -1\)
Step 5: Apply the Chain Rule Formula
Now substitute the partial derivatives calculated in Steps 1-4 into the chain rule formula:
\(\frac{{\partial H}}{{\partial v}} = \frac{{\partial H}}{{\partial x}} \frac{{\partial x}}{{\partial v}} + \frac{{\partial H}}{{\partial y}} \frac{{\partial y}}{{\partial v}}\)
\(\frac{{\partial H}}{{\partial v}} = \left(\frac{y}{x^2 + y^2}\right) \cdot (1) + \left(-\frac{x}{x^2 + y^2}\right) \cdot (-1)\)
\(\frac{{\partial H}}{{\partial v}} = \frac{y}{x^2 + y^2} + \frac{x}{x^2 + y^2}\)
\(\frac{{\partial H}}{{\partial v}} = \frac{x + y}{x^2 + y^2}\)
Step 6: Express the Result in terms of \(u\) and \(v\)
The expression for \(\frac{{\partial H}}{{\partial v}}\) is currently in terms of \(x\) and \(y\). We need to convert it to terms of \(u\) and \(v\) using the given relations \(x = u + v\) and \(y = u - v\).
- Numerator: \(x + y = (u + v) + (u - v) = 2u\)
- Denominator: \(x^2 + y^2 = (u + v)^2 + (u - v)^2\)
- \(x^2 + y^2 = (u^2 + 2uv + v^2) + (u^2 - 2uv + v^2)\)
- \(x^2 + y^2 = u^2 + 2uv + v^2 + u^2 - 2uv + v^2\)
- \(x^2 + y^2 = 2u^2 + 2v^2 = 2(u^2 + v^2)\)
Substitute these back into the expression for \(\frac{{\partial H}}{{\partial v}}\):
\(\frac{{\partial H}}{{\partial v}} = \frac{2u}{2(u^2 + v^2)}\)
\(\frac{{\partial H}}{{\partial v}} = \frac{u}{u^2 + v^2}\)
This final expression is in terms of \(u\) and \(v\).
Comparing this result with the given options, we find that it matches Option 1.
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
x 1 2 3 4 f(x) 20 22 27 35 The set of all points, where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable, is
Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:
If \(f(x)=\displaystyle\sum_{n-0}^{2k}\left(a_n|x|^n+b_n\ \sin^2x\right)\), where \(a_i^{'}\)s and \(b_i^{'}\)s (0 ≤ i ≤ k) are real constants, then f(x) is:
The set of all point where the function f(x) = 2x|x| is differentiable, is:
