Question

problem 1.
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)

Step1: Identify the rule

We use the quotient rule for differentiation. The quotient rule states that if we have a function \( y = \frac{u}{v} \), then \( y'=\frac{u'v - uv'}{v^{2}} \), where \( u = x^{2}\sin(x) \) and \( v=1 + x^{2} \)

Step2: Differentiate \( u \)

To find \( u' \), we use the product rule. The product rule states that if \( u = ab \), then \( u'=a'b+ab' \), where \( a = x^{2} \) and \( b=\sin(x) \)

• Differentiate \( a \): \( a'=\frac{d}{dx}(x^{2}) = 2x \)
• Differentiate \( b \): \( b'=\frac{d}{dx}(\sin(x))=\cos(x) \)
• So, \( u'=2x\sin(x)+x^{2}\cos(x) \)

Step3: Differentiate \( v \)

Differentiate \( v = 1+x^{2} \) with respect to \( x \): \( v'=\frac{d}{dx}(1 + x^{2})=2x \)

Step4: Apply quotient rule

Now, substitute \( u, u', v, v' \) into the quotient rule formula:

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Part (b)

Step1: Identify the rule

We use the product rule and the chain rule. The product rule states that if \( y = uv \), then \( y'=u'v+uv' \), where \( u=\sin^{2}(3x) \) and \( v = \sin(4x^{5}) \)

Step2: Differentiate \( u \)

To differentiate \( u=\sin^{2}(3x) \), we use the chain rule. Let \( t = \sin(3x) \), so \( u=t^{2} \)

• First, differentiate \( u \) with respect to \( t \): \( \frac{du}{dt}=2t \)
• Then, differentiate \( t \) with respect to \( x \): Let \( s = 3x \), so \( t=\sin(s) \)
• Differentiate \( t \) with respect to \( s \): \( \frac{dt}{ds}=\cos(s) \)
• Differentiate \( s \) with respect to \( x \): \( \frac{ds}{dx}=3 \)
• By chain rule, \( \frac{dt}{dx}=\frac{dt}{ds}\cdot\frac{ds}{dx}=3\cos(3x) \)
• Then, by chain rule \( \frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}=2\sin(3x)\cdot3\cos(3x) = 6\sin(3x)\cos(3x) \) (we can also write this as \( 3\sin(6x) \) using the double - angle formula \( \sin(2\theta)=2\sin(\theta)\cos(\theta) \))

Step3: Differentiate \( v \)

To differentiate \( v=\sin(4x^{5}) \), we use the chain rule. Let \( p = 4x^{5} \), so \( v=\sin(p) \)

• Differentiate \( v \) with respect to \( p \): \( \frac{dv}{dp}=\cos(p) \)
• Differentiate \( p \) with respect to \( x \): \( \frac{dp}{dx}=20x^{4} \)
• By chain rule, \( \frac{dv}{dx}=\cos(4x^{5})\cdot20x^{4}=20x^{4}\cos(4x^{5}) \)

Step4: Apply product rule

Now, substitute \( u, u', v, v' \) into the product rule formula:

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Part (c)

Step1: Let's use the chain rule multiple times

Let \( y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}=(1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}} \)
Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y = u^{\frac{1}{2}} \)

• Differentiate \( y \) with respect to \( u \): \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}} \)

Step2: Differentiate \( u \)

Now, \( u = 1 + v \), where \( v=\sqrt{1+\sqrt{1 + x}}=(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \)

• Differentiate \( u \) with respect to \( v \): \( \frac{du}{dv}=1 \)
• Differentiate \( v \) with respect to \( w \), where \( w = 1+\sqrt{1 + x}=(1+(1 + x)^{\frac{1}{2}}) \)
• \( v = w^{\frac{1}{2}} \), so \( \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}}=\frac{1}{2\sqrt{w}}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}} \)
• Differentiate \( w \) with respect to \( z \), where \( z = 1 + x \)
• \( w=1 + z^{\frac{1}{2}} \), so \( \frac{dw}{dz}=\frac{1}{2}z^{-\frac{1}{2}}=\frac{1}{2\sqrt{z}}=\frac{1}{2\sqrt{1 + x}} \)
• Differentiate \( z \) with respect to \( x \): \( \frac{dz}{dx}=1 \)

Step3: Apply chain rule

By the chain rule, \( \frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dw}\cdot\frac{dw}{dz}\cdot\frac{dz}{dx} \)

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Step4: Find \( f'(x) \)

Since \( f'(x)=\frac{dy}{du}\cdot\frac{du}{dx} \)

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Final Answers

(a) \(\boldsymbol{f'(x)=\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}}\) (or the expanded form \(\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}\))

(b) \(\boldsymbol{f'(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})}\) (or the form with \(6\sin(3x)\cos(3x)\))

(c) \(\boldsymbol{f'(x)=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}}\)

Answer:

Step1: Let's use the chain rule multiple times

Let \( y = f(x)=\sqrt{1+\sqrt{1+\sqrt{1 + x}}}=(1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}} \)
Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y = u^{\frac{1}{2}} \)

• Differentiate \( y \) with respect to \( u \): \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}} \)

Step2: Differentiate \( u \)

Now, \( u = 1 + v \), where \( v=\sqrt{1+\sqrt{1 + x}}=(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \)

• Differentiate \( u \) with respect to \( v \): \( \frac{du}{dv}=1 \)
• Differentiate \( v \) with respect to \( w \), where \( w = 1+\sqrt{1 + x}=(1+(1 + x)^{\frac{1}{2}}) \)
• \( v = w^{\frac{1}{2}} \), so \( \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}}=\frac{1}{2\sqrt{w}}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}} \)
• Differentiate \( w \) with respect to \( z \), where \( z = 1 + x \)
• \( w=1 + z^{\frac{1}{2}} \), so \( \frac{dw}{dz}=\frac{1}{2}z^{-\frac{1}{2}}=\frac{1}{2\sqrt{z}}=\frac{1}{2\sqrt{1 + x}} \)
• Differentiate \( z \) with respect to \( x \): \( \frac{dz}{dx}=1 \)

Step3: Apply chain rule

By the chain rule, \( \frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dw}\cdot\frac{dw}{dz}\cdot\frac{dz}{dx} \)

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Step4: Find \( f'(x) \)

Since \( f'(x)=\frac{dy}{du}\cdot\frac{du}{dx} \)

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Final Answers

(a) \(\boldsymbol{f'(x)=\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}}\) (or the expanded form \(\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}\))

(b) \(\boldsymbol{f'(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})}\) (or the form with \(6\sin(3x)\cos(3x)\))

(c) \(\boldsymbol{f'(x)=\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}}\)