Wells Gordon
02/14/2023 · Elementary School
Chain Rule Activity \( y=\frac{\sin (3 t)}{1+t^{2}} \) \( h(u)=\tan (4+10 \mu) \) \( G(x)=2 \sin (3 x+\tan (x)) \) \( g(z)=3 z^{7}-\sin \left(z^{2}+6\right) \)
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Step-by-step Solution
To apply the chain rule to the given functions, we will differentiate each function with respect to its variable. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative is given by:
\[
\frac{dy}{dx} = f'(g(x)) \cdot g'(x)
\]
Let's differentiate each function step by step.
### 1. \( y = \frac{\sin(3t)}{1+t^2} \)
To differentiate this function, we will use the quotient rule, which states that if \( y = \frac{u}{v} \), then:
\[
\frac{dy}{dt} = \frac{u'v - uv'}{v^2}
\]
where \( u = \sin(3t) \) and \( v = 1 + t^2 \).
- \( u' = 3\cos(3t) \) (using the chain rule)
- \( v' = 2t \)
Now applying the quotient rule:
\[
\frac{dy}{dt} = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2}
\]
### 2. \( h(u) = \tan(4 + 10\mu) \)
To differentiate this function, we will use the chain rule. Let \( v = 4 + 10\mu \).
- The derivative of \( h(u) \) with respect to \( u \) is \( h'(u) = \sec^2(v) \cdot \frac{dv}{d\mu} \).
- \( \frac{dv}{d\mu} = 10 \)
Thus,
\[
\frac{dh}{d\mu} = \sec^2(4 + 10\mu) \cdot 10
\]
### 3. \( G(x) = 2\sin(3x + \tan(x)) \)
To differentiate this function, we will use the chain rule. Let \( v = 3x + \tan(x) \).
- The derivative of \( G(x) \) is \( G'(x) = 2\cos(v) \cdot \frac{dv}{dx} \).
- Now, we need to find \( \frac{dv}{dx} = 3 + \sec^2(x) \).
Thus,
\[
G'(x) = 2\cos(3x + \tan(x)) \cdot (3 + \sec^2(x))
\]
### 4. \( g(z) = 3z^7 - \sin(z^2 + 6) \)
To differentiate this function, we will differentiate each term separately.
- The derivative of \( 3z^7 \) is \( 21z^6 \).
- For \( -\sin(z^2 + 6) \), we use the chain rule:
- Let \( v = z^2 + 6 \), then \( \frac{dv}{dz} = 2z \).
- The derivative of \( -\sin(v) \) is \( -\cos(v) \cdot \frac{dv}{dz} = -\cos(z^2 + 6) \cdot 2z \).
Putting it all together:
\[
g'(z) = 21z^6 - 2z\cos(z^2 + 6)
\]
### Summary of Derivatives
1. \( \frac{dy}{dt} = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2} \)
2. \( \frac{dh}{d\mu} = 10\sec^2(4 + 10\mu) \)
3. \( G'(x) = 2\cos(3x + \tan(x))(3 + \sec^2(x)) \)
4. \( g'(z) = 21z^6 - 2z\cos(z^2 + 6) \)
Quick Answer
1. \( \frac{dy}{dt} = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2} \)
2. \( \frac{dh}{d\mu} = 10\sec^2(4 + 10\mu) \)
3. \( G'(x) = 2\cos(3x + \tan(x))(3 + \sec^2(x)) \)
4. \( g'(z) = 21z^6 - 2z\cos(z^2 + 6) \)
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