Question

differentiate the following functions.
(a) 4pts. $f(x) = \frac{x^2 \sin(x)}{1 + x^2}$
(b) 4pts. $f(x) = \sin^2(3x) \sin(4x^5)$
(c) 4pts. $f(x) = \sqrt{1 + \sqrt{1 + \sqrt{1 + x}}}$

Explanation:

Part (a)

Step 1: Identify the quotient rule

We have a function \( f(x) = \frac{u(x)}{v(x)} \) where \( u(x) = x^2\sin(x) \) and \( v(x) = 1 + x^2 \). The quotient rule states that \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

Step 2: Find \( u'(x) \)

To find \( u'(x) \), we use the product rule. If \( u(x) = x^2\sin(x) \), let \( a(x) = x^2 \) and \( b(x) = \sin(x) \). The product rule is \( (a(x)b(x))' = a'(x)b(x) + a(x)b'(x) \).

• \( a'(x) = 2x \)
• \( b'(x) = \cos(x) \)

So, \( u'(x) = 2x\sin(x) + x^2\cos(x) \)

Step 3: Find \( v'(x) \)

For \( v(x) = 1 + x^2 \), the derivative \( v'(x) = 2x \)

Step 4: Apply the quotient rule

Substitute \( u(x) \), \( u'(x) \), \( v(x) \), and \( v'(x) \) into the quotient rule formula:

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Step 1: Identify the product rule

We have a function \( f(x) = g(x)h(x) \) where \( g(x) = \sin^2(3x) \) and \( h(x) = \sin(4x^5) \). The product rule states that \( f'(x) = g'(x)h(x) + g(x)h'(x) \).

Step 2: Find \( g'(x) \)

To find \( g'(x) \), we use the chain rule. Let \( u = \sin(3x) \), so \( g(x) = u^2 \). The chain rule states that \( \frac{d}{dx}[g(x)] = \frac{d}{du}[u^2] \cdot \frac{du}{dx} \).

• \( \frac{d}{du}[u^2] = 2u \)
• To find \( \frac{du}{dx} \), let \( v = 3x \), so \( u = \sin(v) \). Then \( \frac{du}{dx} = \cos(v) \cdot \frac{dv}{dx} = \cos(3x) \cdot 3 = 3\cos(3x) \)

So, \( g'(x) = 2\sin(3x) \cdot 3\cos(3x) = 6\sin(3x)\cos(3x) \) (we can also write this as \( 3\sin(6x) \) using the double-angle identity, but we'll keep it as \( 6\sin(3x)\cos(3x) \) for now)

Step 3: Find \( h'(x) \)

To find \( h'(x) \), we use the chain rule. Let \( w = 4x^5 \), so \( h(x) = \sin(w) \). The chain rule states that \( \frac{d}{dx}[h(x)] = \cos(w) \cdot \frac{dw}{dx} \).

• \( \frac{dw}{dx} = 20x^4 \)

So, \( h'(x) = \cos(4x^5) \cdot 20x^4 = 20x^4\cos(4x^5) \)

Step 4: Apply the product rule

Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \) into the product rule formula:

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Or, using the double-angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \), we can rewrite \( 6\sin(3x)\cos(3x) = 3\sin(6x) \), so:
\( f'(x) = 3\sin(6x)\sin(4x^5) + 20x^4\sin^2(3x)\cos(4x^5) \)

Step 1: Use the chain rule multiple times

Let's define the function layer by layer. Let \( y = f(x) = \sqrt{1 + \sqrt{1 + \sqrt{1 + x}}} \). Let \( u = 1 + \sqrt{1 + \sqrt{1 + x}} \), so \( y = \sqrt{u} = u^{1/2} \). The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

Step 2: Find \( \frac{dy}{du} \)

\( \frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}} \)

Step 3: Find \( \frac{du}{dx} \)

\( u = 1 + v \) where \( v = \sqrt{1 + \sqrt{1 + x}} \). So, \( \frac{du}{dx} = \frac{dv}{dx} \) (since the derivative of 1 is 0). Now, let's find \( \frac{dv}{dx} \). Let \( w = 1 + \sqrt{1 + x} \), so \( v = \sqrt{w} = w^{1/2} \). Then \( \frac{dv}{dx} = \frac{1}{2}w^{-1/2} \cdot \frac{dw}{dx} \) (by the chain rule).

Step 4: Find \( \frac{dw}{dx} \)

\( w = 1 + z \) where \( z = \sqrt{1 + x} \). So, \( \frac{dw}{dx} = \frac{dz}{dx} \) (since the derivative of 1 is 0). Now, let's find \( \frac{dz}{dx} \). Let \( t = 1 + x \), so \( z = \sqrt{t} = t^{1/2} \). Then \( \frac{dz}{dx} = \frac{1}{2}t^{-1/2} \cdot \frac{dt}{dx} \) (by the chain rule).

Step 5: Find \( \frac{dt}{dx} \)

\( t = 1 + x \), so \( \frac{dt}{dx} = 1 \)

Step 6: Work backwards to find \( \frac{dz}{dx} \)

Substitute \( \frac{dt}{dx} = 1 \) into the expression for \( \frac{dz}{dx} \):
\( \frac{dz}{dx} = \frac{1}{2}(1 + x)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{1 + x}} \)

Step 7: Work backwards to find \( \frac{dv}{dx} \)

Substitute \( \frac{dz}{dx} \) into the expression for \( \frac{dv}{dx} \). First, \( w = 1 + z = 1 + \sqrt{1 + x} \), so:
\( \frac{dv}{dx} = \frac{1}{2}(1 + \sqrt{1 + x})^{-1/2} \cdot \frac{1}{2\sqrt{1 + x}} = \frac{1}{4\sqrt{1 + \sqrt{1 + x}} \cdot \sqrt{1 + x}} \)

Step 8: Work backwards to find \( \frac{du}{dx} \)

Since \( \frac{du}{dx} = \frac{dv}{dx} \), we have:
\( \frac{du}{dx} = \frac{1}{4\sqrt{1 + \sqrt{1 + x}} \cdot \sqrt{1 + x}} \)

Step 9: Work backwards to find \( \frac{dy}{dx} \)

Substitute \( \frac{du}{dx} \) into the expression for \( \frac{dy}{dx} \). Recall that \( u = 1 + \sqrt{1 + \sqrt{1 + x}} \), so:

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Answer:

\( f'(x) = \frac{2x\sin(x) + x^2(1 + x^2)\cos(x)}{(1 + x^2)^2} \)

Part (b)