Answer :
Let's find the derivatives of the given functions step by step.
### Part (a): Derivative of [tex]\( x^{\frac{\cosh x}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\cosh x}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\cosh x}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\cosh x}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\cosh x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\cosh x \ln x)}{\partial x} = \cosh x \cdot \frac{1}{x} + \sinh x \cdot \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\cosh x}{x} + \sinh x \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\cosh x}{a}} \left( \frac{\cosh x}{a x} + \frac{\sinh x \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\cosh x}{a}} \right) = x^{\frac{\cosh x}{a}} \left( \frac{\cosh x}{a x} + \frac{\sinh x \ln x}{a} \right). \][/tex]
### Part (b): Derivative of [tex]\( x^{\frac{\sinh x^2}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\sinh x^2}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\sinh x^2}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\sinh x^2}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\sinh x^2 \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\sinh x^2 \ln x)}{\partial x} = \sinh x^2 \cdot \frac{1}{x} + \cosh x^2 \cdot 2x \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\sinh x^2}{x} + 2x \cosh x^2 \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\sinh x^2}{a}} \left( \frac{\sinh x^2}{a x} + \frac{2x \cosh x^2 \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\sinh x^2}{a}} \right) = x^{\frac{\sinh x^2}{a}} \left( \frac{\sinh x^2}{a x} + \frac{2x \cosh x^2 \ln x}{a} \right). \][/tex]
### Part (c): Derivative of [tex]\( x^{\frac{\cosh^2 x}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\cosh^2 x}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\cosh^2 x}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\cosh^2 x}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\cosh^2 x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\cosh^2 x \ln x)}{\partial x} = \cosh^2 x \cdot \frac{1}{x} + 2 \cosh x \sinh x \cdot \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\cosh^2 x}{x} + 2 \cosh x \sinh x \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\cosh^2 x}{a}} \left( \frac{\cosh^2 x}{a x} + \frac{2 \cosh x \sinh x \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\cosh^2 x}{a}} \right) = x^{\frac{\cosh^2 x}{a}} \left( \frac{\cosh^2 x}{a x} + 2 \cosh x \sinh x \ln x \right). \][/tex]
### Part (d): Derivative of [tex]\( x^{\frac{\tanh^{-1} x}{3}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\tanh^{-1} x}{3}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\tanh^{-1} x}{3}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\tanh^{-1} x}{3} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{\partial (\tanh^{-1} x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\tanh^{-1} x \ln x)}{\partial x} = \tanh^{-1} x \cdot \frac{1}{x} + \frac{\ln x}{1 - x^2} \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{\tanh^{-1} x}{x} + \frac{\ln x}{1 - x^2} \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\tanh^{-1} x}{3}} \left( \frac{\tanh^{-1} x}{3 x} + \frac{\ln x}{3 (1 - x^2)} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\tanh^{-1} x}{3}} \right) = x^{\frac{\tanh^{-1} x}{3}} \left( \frac{\tanh^{-1} x}{3 x} + \frac{\ln x}{3 (1 - x^2)} \right). \][/tex]
### Part (a): Derivative of [tex]\( x^{\frac{\cosh x}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\cosh x}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\cosh x}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\cosh x}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\cosh x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\cosh x \ln x)}{\partial x} = \cosh x \cdot \frac{1}{x} + \sinh x \cdot \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\cosh x}{x} + \sinh x \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\cosh x}{a}} \left( \frac{\cosh x}{a x} + \frac{\sinh x \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\cosh x}{a}} \right) = x^{\frac{\cosh x}{a}} \left( \frac{\cosh x}{a x} + \frac{\sinh x \ln x}{a} \right). \][/tex]
### Part (b): Derivative of [tex]\( x^{\frac{\sinh x^2}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\sinh x^2}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\sinh x^2}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\sinh x^2}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\sinh x^2 \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\sinh x^2 \ln x)}{\partial x} = \sinh x^2 \cdot \frac{1}{x} + \cosh x^2 \cdot 2x \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\sinh x^2}{x} + 2x \cosh x^2 \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\sinh x^2}{a}} \left( \frac{\sinh x^2}{a x} + \frac{2x \cosh x^2 \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\sinh x^2}{a}} \right) = x^{\frac{\sinh x^2}{a}} \left( \frac{\sinh x^2}{a x} + \frac{2x \cosh x^2 \ln x}{a} \right). \][/tex]
### Part (c): Derivative of [tex]\( x^{\frac{\cosh^2 x}{a}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\cosh^2 x}{a}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\cosh^2 x}{a}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\cosh^2 x}{a} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{a} \left( \frac{\partial (\cosh^2 x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\cosh^2 x \ln x)}{\partial x} = \cosh^2 x \cdot \frac{1}{x} + 2 \cosh x \sinh x \cdot \ln x \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{a} \left( \frac{\cosh^2 x}{x} + 2 \cosh x \sinh x \ln x \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\cosh^2 x}{a}} \left( \frac{\cosh^2 x}{a x} + \frac{2 \cosh x \sinh x \ln x}{a} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\cosh^2 x}{a}} \right) = x^{\frac{\cosh^2 x}{a}} \left( \frac{\cosh^2 x}{a x} + 2 \cosh x \sinh x \ln x \right). \][/tex]
### Part (d): Derivative of [tex]\( x^{\frac{\tanh^{-1} x}{3}} \)[/tex]
1. Given function: [tex]\( f(x) = x^{\frac{\tanh^{-1} x}{3}} \)[/tex]
2. Rewrite using natural logarithm: Let [tex]\( y = x^{\frac{\tanh^{-1} x}{3}} \)[/tex]. Taking the natural logarithm on both sides, we have:
[tex]\[ \ln y = \frac{\tanh^{-1} x}{3} \ln x \][/tex]
3. Differentiate implicitly with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \left( \frac{\partial (\tanh^{-1} x \ln x)}{\partial x} \right) \][/tex]
Using the product rule:
[tex]\[ \frac{\partial (\tanh^{-1} x \ln x)}{\partial x} = \tanh^{-1} x \cdot \frac{1}{x} + \frac{\ln x}{1 - x^2} \][/tex]
Therefore,
[tex]\[ \frac{dy}{dx} = y \cdot \frac{1}{3} \left( \frac{\tanh^{-1} x}{x} + \frac{\ln x}{1 - x^2} \right) \][/tex]
4. Substitute [tex]\( y \)[/tex] back:
[tex]\[ \frac{dy}{dx} = x^{\frac{\tanh^{-1} x}{3}} \left( \frac{\tanh^{-1} x}{3 x} + \frac{\ln x}{3 (1 - x^2)} \right) \][/tex]
Thus, the derivative is:
[tex]\[ \frac{d}{dx} \left( x^{\frac{\tanh^{-1} x}{3}} \right) = x^{\frac{\tanh^{-1} x}{3}} \left( \frac{\tanh^{-1} x}{3 x} + \frac{\ln x}{3 (1 - x^2)} \right). \][/tex]
