Question

Berechnen Sie die Ableitungen der Funktion $$ f_{1}(x)=\sqrt{\sin x}, \quad f_{2}(x)=x^{x^{2}}, \quad f_{3}(x)=\frac{x e^{x}}{\arctan x}, \quad f_{4}(x)=x \sqrt{x} \cos x $$

Short answer

The derivatives are \( f_1'(x) = \frac{\cos x}{2\sqrt{\sin x}} \), \( f_2'(x) = x^{x^2}(2x \ln x + x) \), \( f_3'(x) = \frac{(e^x + xe^x)\arctan x - xe^x \cdot \frac{1}{1+x^2}}{(\arctan x)^2} \), \( f_4'(x) = \frac{3}{2}x^{1/2}\cos x - x^{3/2}\sin x \).

Step-by-step solution

Differentiate f₁(x) = √(sin x)

To find the derivative of \( f_1(x) = \sqrt{\sin x} \), use the chain rule. Let \( u = \sin x \), so \( f_1(x) = u^{1/2} \). The derivative of \( u^{1/2} \) with respect to \( u \) is \( \frac{1}{2}u^{-1/2} \), and the derivative of \( u = \sin x \) with respect to \( x \) is \( \cos x \). Therefore, the derivative \( f_1'(x) = \frac{1}{2}(\sin x)^{-1/2} \cdot \cos x = \frac{\cos x}{2\sqrt{\sin x}} \).

Differentiate f₂(x) = x^{x²}

For the function \( f_2(x) = x^{x^2} \), use logarithmic differentiation. Let \( y = x^{x^2} \). Taking the natural logarithm on both sides gives \( \ln y = x^2 \ln x \). Differentiate implicitly with respect to \( x \): \( \frac{1}{y} \frac{dy}{dx} = 2x\ln x + x^2 \cdot \frac{1}{x} \). Thus, \( \frac{dy}{dx} = y \left( 2x\ln x + x \right) \). Substitute back \( y = x^{x^2} \) to get \( f_2'(x) = x^{x^2} \left( 2x \ln x + x \right) \).

Differentiate f₃(x) = \( \frac{x e^x}{\arctan x} \)

To differentiate \( f_3(x) = \frac{x e^x}{\arctan x} \), use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \). Let \( u = xe^x \) and \( v = \arctan x \). Then, \( u' = e^x + xe^x \) and \( v' = \frac{1}{1+x^2} \). Applying the quotient rule: \( f_3'(x) = \frac{(e^x + xe^x)\arctan x - xe^x \left(\frac{1}{1+x^2}\right)}{(\arctan x)^2} \).

Differentiate f₄(x) = x√x cos x

For \( f_4(x) = x\sqrt{x}\cos x \), first rewrite as \( f_4(x) = x^{3/2}\cos x \). Use the product rule: \( (uv)' = u'v + uv' \). Let \( u = x^{3/2} \) and \( v = \cos x \). Then, \( u' = \frac{3}{2}x^{1/2} \) and \( v' = -\sin x \). Apply the product rule: \( f_4'(x) = \frac{3}{2}x^{1/2}\cos x - x^{3/2}\sin x \).

Key concepts

Chain Rule

When it comes to calculus differentiation, the chain rule is a powerful tool used to differentiate composite functions. A composite function is essentially a function within another function. The chain rule states that the derivative of a composite function \( f(g(x)) \) is the derivative of \( f \) with respect to \( g \), multiplied by the derivative of \( g \) with respect to \( x \). This can be expressed as:\[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]In simpler terms, you "chain" together the derivatives of the inner and outer functions. For example, if you have \( f_1(x) = \sqrt{\sin x} \), let \( u = \sin x \), so \( f_1(x) = u^{1/2} \). The derivative becomes the product of the derivative of \( u^{1/2} \) with respect to \( u \), which is \( \frac{1}{2}u^{-1/2} \), and the derivative of \( \sin x \) with respect to \( x \), which is \( \cos x \). Therefore, the chain rule helps you find the derivative of the composite function as \( \frac{\cos x}{2\sqrt{\sin x}} \).

It's crucial to get the right order: start from the outer function and move inward. Don't forget to multiply by the derivative of each inner layer.

Quotient Rule

The quotient rule is employed when you need to differentiate a function defined as the quotient of two other functions, expressed as \( \frac{u}{v} \). The quotient rule provides a systematic method to find the derivative, defined by:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]In essence, this rule states that you need to find the derivative of the numerator \( (u') \) times the denominator \( (v) \), subtract the product of the numerator \( (u) \) and the derivative of the denominator \( (v') \), and divide it all by the square of the denominator \( (v^2) \).

Let's see this in action with the function \( f_3(x) = \frac{x e^x}{\arctan x} \). We define \( u = x e^x \) and \( v = \arctan x \). Hence, \( u' = e^x + x e^x \) and \( v' = \frac{1}{1+x^2} \). Substituting these into the quotient rule formula gives the derivative for \( f_3(x) \).

Ensure all your steps are clear and organized, as mixing terms or forgetting a part of the formula will lead to incorrect answers.

Product Rule

The product rule is applicable when differentiating products of two functions. It's defined by the formula:\[(uv)' = u'v + uv'\]Simply put, you take the derivative of the first function and multiply it by the second, then add the product of the first function and the derivative of the second function.

For instance, when we look at the function \( f_4(x) = x\sqrt{x}\cos x \), it's helpful to first rewrite it as \( (x^{3/2})(\cos x) \). This gives a clearer structure to apply the product rule. Let \( u = x^{3/2} \) and \( v = \cos x \). Thus, the derivatives are \( u' = \frac{3}{2}x^{1/2} \) and \( v' = -\sin x \). Plug these into the product rule formula to find the derivative.

The product rule ensures that you account for both functions' contributions to the rate of change, providing an accurate derivative for complex products.

Logarithmic Differentiation

Logarithmic differentiation is a technique especially handy for functions that involve variables in both the base and the exponent, such as \( f_2(x) = x^{x^2} \). This method involves three main steps:

• Take the natural logarithm of both sides of the given function.
• Differentiate implicitly with respect to \( x \).
• Solve for the derivative, and substitute back the original function.

Let's see this with \( f_2(x) = x^{x^2} \). Start by setting \( y = x^{x^2} \) and take the natural log: \( \ln y = x^2 \ln x \). Differentiate the equation implicitly: \( \frac{1}{y} \frac{dy}{dx} = 2x\ln x + x^2 \cdot \frac{1}{x} \). Finally, solve for \( \frac{dy}{dx} \) by multiplying by \( y \) and substitute \( y = x^{x^2} \) into the equation.

This technique simplifies differentiating complex forms by breaking them down into more manageable steps using properties of logarithms, making it easier to handle otherwise cumbersome derivatives.