Question

Find the differential coefficient of \(y = x^{\cos x} + (\sin x)^x\) with respect to x.

Hint:

Whenever you encounter a function in the form of (variable)\textsuperscript{(variable)}, i.e., \(f(x)^{g(x)}\), your first thought should be logarithmic differentiation. Remember that \(\frac{d}{dx}(u+v) = \frac{du}{dx} + \frac{dv}{dx}\), but you cannot apply logarithms directly to a sum like \(\ln(u+v)\). Always handle each term separately.

Answer 1

Step 1: Understanding the Concept:
The function is a sum of two functions, each of the form \(f(x)^{g(x)}\).
This type of function requires logarithmic differentiation. We cannot differentiate it directly using standard power or exponential rules.
We will differentiate each term separately using the sum rule of differentiation.
Step 2: Key Formula or Approach:
1. Let \(y = u(x) + v(x)\), where \(u(x) = x^{\cos x}\) and \(v(x) = (\sin x)^x\).
2. The derivative is \(\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}\).
3. For functions of the form \(f(x)^{g(x)}\), apply logarithms: take the natural log of both sides, use log properties to bring the exponent down, and then differentiate implicitly.
If \(u = f^g\), then \(\ln u = g \ln f\), and \(\frac{1}{u}\frac{du}{dx} = g' \ln f + g \frac{f'}{f}\).
Step 3: Detailed Explanation or Calculation:
Part 1: Differentiating \(u = x^{\cos x}\)
Take the natural logarithm of both sides:
\[ \ln u = \ln(x^{\cos x}) \] \[ \ln u = \cos x \ln x \] Differentiate both sides with respect to x using the product rule:
\[ \frac{1}{u} \frac{du}{dx} = (-\sin x)(\ln x) + (\cos x)\left(\frac{1}{x}\right) \] \[ \frac{du}{dx} = u \left( \frac{\cos x}{x} - \sin x \ln x \right) \] Substitute back \(u = x^{\cos x}\):
\[ \frac{du}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right) \] Part 2: Differentiating \(v = (\sin x)^x\)
Take the natural logarithm of both sides:
\[ \ln v = \ln((\sin x)^x) \] \[ \ln v = x \ln(\sin x) \] Differentiate both sides with respect to x using the product rule:
\[ \frac{1}{v} \frac{dv}{dx} = (1)(\ln(\sin x)) + x \left(\frac{1}{\sin x} \cdot \cos x\right) \] \[ \frac{1}{v} \frac{dv}{dx} = \ln(\sin x) + x \cot x \] \[ \frac{dv}{dx} = v (\ln(\sin x) + x \cot x) \] Substitute back \(v = (\sin x)^x\):
\[ \frac{dv}{dx} = (\sin x)^x (\ln(\sin x) + x \cot x) \] Part 3: Combining the results
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \] \[ \frac{dy}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right) + (\sin x)^x (\ln(\sin x) + x \cot x) \] Step 4: Final Answer
The differential coefficient is \(x^{\cos x} \left( \frac{\cos x}{x} - \sin x \ln x \right) + (\sin x)^x (\ln(\sin x) + x \cot x)\).