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)

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

Step 1: Find \( u'(x) \)

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

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

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

Step 3: Apply the quotient rule

\( f'(x)=\frac{(2x\sin(x)+x^{2}\cos(x))(1 + x^{2})-x^{2}\sin(x)\times(2x)}{(1 + x^{2})^{2}} \)
Simplify the numerator:

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So, \( f'(x)=\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}=\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}} \)

Part (b)

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

Step 1: Find \( u'(x) \) for \( u(x)=\sin^{2}(3x) \)

Let \( t = \sin(3x) \), so \( u=t^{2} \). Then \( \frac{du}{dt}=2t \) and \( \frac{dt}{dx}=3\cos(3x) \) (by chain rule, derivative of \( \sin(3x) \) is \( 3\cos(3x) \)). So,
\( u'(x)=2\sin(3x)\times3\cos(3x)=6\sin(3x)\cos(3x) = 3\sin(6x) \) (using the double - angle formula \( \sin(2\theta)=2\sin\theta\cos\theta \), here \( \theta = 3x \), so \( 2\sin(3x)\cos(3x)=\sin(6x) \))

Step 2: Find \( v'(x) \) for \( v(x)=\sin(4x^{5}) \)

Let \( t = 4x^{5} \), so \( v=\sin(t) \). Then \( \frac{dv}{dt}=\cos(t) \) and \( \frac{dt}{dx}=20x^{4} \). So,
\( v'(x)=\cos(4x^{5})\times20x^{4}=20x^{4}\cos(4x^{5}) \)

Step 3: Apply the product rule

\( f'(x)=u'(x)v(x)+u(x)v'(x) \)
\( f'(x)=3\sin(6x)\sin(4x^{5})+\sin^{2}(3x)\times20x^{4}\cos(4x^{5}) \)
\( f'(x)=3\sin(6x)\sin(4x^{5})+20x^{4}\sin^{2}(3x)\cos(4x^{5}) \)

Part (c)

We use the chain rule multiple times. Let \( y=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} \). Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y=\sqrt{u}=u^{\frac{1}{2}} \). Then we find derivatives step - by - step.

Step 1: Find \( \frac{dy}{du} \)

If \( y = u^{\frac{1}{2}} \), then \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}} \)

Step 2: Find \( \frac{du}{dv} \) where \( v = 1+\sqrt{1+\sqrt{1 + x}} \)

Let \( w=\sqrt{1+\sqrt{1 + x}} \), so \( v = 1 + w \). Then \( \frac{dv}{dw}=1 \) and we need to find \( \frac{dw}{dx} \)

Step 3: Find \( \frac{dw}{dx} \) where \( w=\sqrt{1+\sqrt{1 + x}} \)

Let \( z=\sqrt{1 + x} \), so \( w=\sqrt{1 + z}=(1 + z)^{\frac{1}{2}} \). Then \( \frac{dw}{dz}=\frac{1}{2}(1 + z)^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 + z}}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}} \) and \( \frac{dz}{dx}=\frac{1}{2\sqrt{1 + x}} \) (since derivative of \( \sqrt{1 + x}=(1 + x)^{\frac{1}{2}} \) is \( \frac{1}{2}(1 + x)^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 + x}} \))

By chain rule, \( \frac{dw}{dx}=\frac{dw}{dz}\times\frac{dz}{dx}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\times\frac{1}{2\sqrt{1 + x}}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Then \( \frac{dv}{dx}=\frac{dv}{dw}\times\frac{dw}{dx}=1\times\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Step 4: Find \( \frac{du}{dx} \) (since \( u = 1 + v \))

\( \frac{du}{dx}=\frac{dv}{dx}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Step 5: Apply the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \)

\( \frac{dy}{dx}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\times\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)
Simplify the denominator:
\( 2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\times4\sqrt{(1+\sqrt{1 + x})(1 + x)} = 8\sqrt{(1+\sqrt{1+\sqrt{1 + x}})(1+\sqrt{1 + x})(1 + x)} \)
So, \( f'(x)=\frac{1}{8\sqrt{(1+\sqrt{1+\sqrt{1 + x}})(1+\sqrt{1 + x})(1 + x)}} \)

Part (a) Answer: \(\boldsymbol{\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}}\) (or simplified form \(\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}\))

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

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

Answer:

We use the chain rule multiple times. Let \( y=\sqrt{1+\sqrt{1+\sqrt{1 + x}}} \). Let \( u = 1+\sqrt{1+\sqrt{1 + x}} \), so \( y=\sqrt{u}=u^{\frac{1}{2}} \). Then we find derivatives step - by - step.

Step 1: Find \( \frac{dy}{du} \)

If \( y = u^{\frac{1}{2}} \), then \( \frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}} \)

Step 2: Find \( \frac{du}{dv} \) where \( v = 1+\sqrt{1+\sqrt{1 + x}} \)

Let \( w=\sqrt{1+\sqrt{1 + x}} \), so \( v = 1 + w \). Then \( \frac{dv}{dw}=1 \) and we need to find \( \frac{dw}{dx} \)

Step 3: Find \( \frac{dw}{dx} \) where \( w=\sqrt{1+\sqrt{1 + x}} \)

Let \( z=\sqrt{1 + x} \), so \( w=\sqrt{1 + z}=(1 + z)^{\frac{1}{2}} \). Then \( \frac{dw}{dz}=\frac{1}{2}(1 + z)^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 + z}}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}} \) and \( \frac{dz}{dx}=\frac{1}{2\sqrt{1 + x}} \) (since derivative of \( \sqrt{1 + x}=(1 + x)^{\frac{1}{2}} \) is \( \frac{1}{2}(1 + x)^{-\frac{1}{2}}=\frac{1}{2\sqrt{1 + x}} \))

By chain rule, \( \frac{dw}{dx}=\frac{dw}{dz}\times\frac{dz}{dx}=\frac{1}{2\sqrt{1+\sqrt{1 + x}}}\times\frac{1}{2\sqrt{1 + x}}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Then \( \frac{dv}{dx}=\frac{dv}{dw}\times\frac{dw}{dx}=1\times\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Step 4: Find \( \frac{du}{dx} \) (since \( u = 1 + v \))

\( \frac{du}{dx}=\frac{dv}{dx}=\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)

Step 5: Apply the chain rule \( \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx} \)

\( \frac{dy}{dx}=\frac{1}{2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}}\times\frac{1}{4\sqrt{(1+\sqrt{1 + x})(1 + x)}} \)
Simplify the denominator:
\( 2\sqrt{1+\sqrt{1+\sqrt{1 + x}}}\times4\sqrt{(1+\sqrt{1 + x})(1 + x)} = 8\sqrt{(1+\sqrt{1+\sqrt{1 + x}})(1+\sqrt{1 + x})(1 + x)} \)
So, \( f'(x)=\frac{1}{8\sqrt{(1+\sqrt{1+\sqrt{1 + x}})(1+\sqrt{1 + x})(1 + x)}} \)

Part (a) Answer: \(\boldsymbol{\frac{2x\sin(x)+x^{2}\cos(x)+x^{4}\cos(x)}{(1 + x^{2})^{2}}}\) (or simplified form \(\frac{x(2\sin(x)+x\cos(x)+x^{3}\cos(x))}{(1 + x^{2})^{2}}\))

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

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