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)

Step 1: Identify the rule

We use the quotient rule for differentiation, which states that if \( f(x)=\frac{u(x)}{v(x)} \), then \( f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{[v(x)]^{2}} \). Here, \( u(x) = x^{2}\sin(x) \) and \( v(x)=1 + x^{2} \).

Step 2: Differentiate \( u(x) \)

We use the product rule for \( u(x)=x^{2}\sin(x) \). The product rule states that if \( u(x)=a(x)b(x) \), then \( u^{\prime}(x)=a^{\prime}(x)b(x)+a(x)b^{\prime}(x) \). Let \( a(x)=x^{2} \) and \( b(x)=\sin(x) \). Then \( a^{\prime}(x) = 2x \) and \( b^{\prime}(x)=\cos(x) \). So, \( u^{\prime}(x)=2x\sin(x)+x^{2}\cos(x) \).

Step 3: Differentiate \( v(x) \)

For \( v(x)=1 + x^{2} \), the derivative \( v^{\prime}(x)=2x \).

Step 4: Apply the quotient rule

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Part (b)

Step 1: Identify the rule

We use the product rule and the chain rule. The product rule: if \( f(x)=u(x)v(x) \), then \( f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x) \). Let \( u(x)=\sin^{2}(3x) \) and \( v(x)=\sin(4x^{5}) \).

Step 2: Differentiate \( u(x) \)

Using the chain rule. Let \( y = \sin^{2}(3x)=(\sin(3x))^{2} \). Let \( u=\sin(3x) \), then \( y = u^{2} \). \( \frac{dy}{du}=2u \) and \( \frac{du}{dx}=3\cos(3x) \). So, \( \frac{dy}{dx}=2\sin(3x)\times3\cos(3x)=6\sin(3x)\cos(3x)=3\sin(6x) \) (using the double - angle formula \( \sin(2\theta)=2\sin\theta\cos\theta \)).

Step 3: Differentiate \( v(x) \)

Using the chain rule. Let \( y=\sin(4x^{5}) \). Let \( u = 4x^{5} \), then \( y=\sin(u) \). \( \frac{dy}{du}=\cos(u) \) and \( \frac{du}{dx}=20x^{4} \). So, \( \frac{dy}{dx}=\cos(4x^{5})\times20x^{4}=20x^{4}\cos(4x^{5}) \).

Step 4: Apply the product rule

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Part (c)

Step 1: 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 \( u = 1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \), so \( y = u^{\frac{1}{2}} \), and \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}} \)

Step 2: Differentiate \( u \)

Let \( v=1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \), so \( u = 1 + v^{\frac{1}{2}} \), \( \frac{du}{dv}=\frac{1}{2}v^{-\frac{1}{2}} \)

Step 3: Differentiate \( v \)

Let \( w = 1+(1 + x)^{\frac{1}{2}} \), so \( v = 1+w^{\frac{1}{2}} \), \( \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}} \)

Step 4: Differentiate \( w \)

Let \( z=1 + x \), so \( w = 1+z^{\frac{1}{2}} \), \( \frac{dw}{dz}=\frac{1}{2}z^{-\frac{1}{2}} \) and \( \frac{dz}{dx}=1 \)

Step 5: Multiply the derivatives (chain rule)

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Part (a) Answer: \(\boldsymbol{\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}}\)

Part (b) Answer: \(\boldsymbol{3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})}\)

Part (c) Answer: \(\boldsymbol{\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}}\)

Answer:

Step 1: 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 \( u = 1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \), so \( y = u^{\frac{1}{2}} \), and \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}} \)

Step 2: Differentiate \( u \)

Let \( v=1+(1+(1 + x)^{\frac{1}{2}})^{\frac{1}{2}} \), so \( u = 1 + v^{\frac{1}{2}} \), \( \frac{du}{dv}=\frac{1}{2}v^{-\frac{1}{2}} \)

Step 3: Differentiate \( v \)

Let \( w = 1+(1 + x)^{\frac{1}{2}} \), so \( v = 1+w^{\frac{1}{2}} \), \( \frac{dv}{dw}=\frac{1}{2}w^{-\frac{1}{2}} \)

Step 4: Differentiate \( w \)

Let \( z=1 + x \), so \( w = 1+z^{\frac{1}{2}} \), \( \frac{dw}{dz}=\frac{1}{2}z^{-\frac{1}{2}} \) and \( \frac{dz}{dx}=1 \)

Step 5: Multiply the derivatives (chain rule)

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Part (a) Answer: \(\boldsymbol{\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}}\)

Part (b) Answer: \(\boldsymbol{3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5})}\)

Part (c) Answer: \(\boldsymbol{\frac{1}{8\sqrt{1 + x}\sqrt{1+\sqrt{1 + x}}\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}}\)