Dunn Sanders
11/30/2023 · Middle School
Chain Rule Activity 1.) \( y=\frac{\sin (3 t)}{1+t^{2}} \) 2.) \( h(u)=\tan (4+10 \mu) \) 3.) \( G(x)=2 \sin (3 x+\tan (x)) \) 4.) \( g(z)=3 z^{7}-\sin \left(z^{2}+6\right) \)
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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 \( f(g(x)) \), then the derivative is given by \( f'(g(x)) \cdot g'(x) \).
Let's differentiate each function step by step.
### 1.) \( y=\frac{\sin (3 t)}{1+t^{2}} \)
To differentiate this function, we will use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \).
Let:
- \( u = \sin(3t) \)
- \( v = 1 + t^2 \)
Now, we find \( u' \) and \( v' \):
- \( u' = \cos(3t) \cdot 3 = 3\cos(3t) \) (using the chain rule)
- \( v' = 2t \)
Now apply the quotient rule:
\[
y' = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2}
\]
### 2.) \( h(u)=\tan (4+10 \mu) \)
To differentiate this function, we will use the chain rule.
Let:
- \( g(\mu) = 4 + 10\mu \)
- \( h(g) = \tan(g) \)
Now, we find \( g' \) and \( h' \):
- \( g' = 10 \)
- \( h' = \sec^2(g) \)
Now apply the chain rule:
\[
h' = h'(g) \cdot g' = \sec^2(4 + 10\mu) \cdot 10
\]
So,
\[
h'(\mu) = 10 \sec^2(4 + 10\mu)
\]
### 3.) \( G(x)=2 \sin (3 x+\tan (x)) \)
To differentiate this function, we will use the chain rule.
Let:
- \( g(x) = 3x + \tan(x) \)
- \( G(g) = 2\sin(g) \)
Now, we find \( g' \) and \( G' \):
- \( g' = 3 + \sec^2(x) \)
- \( G' = 2\cos(g) \)
Now apply the chain rule:
\[
G'(x) = G'(g) \cdot g' = 2\cos(3x + \tan(x)) \cdot (3 + \sec^2(x))
\]
### 4.) \( g(z)=3 z^{7}-\sin \left(z^{2}+6\right) \)
To differentiate this function, we will differentiate each term separately.
The derivative of the first term:
\[
\frac{d}{dz}(3z^7) = 21z^6
\]
For the second term, we will use the chain rule:
Let:
- \( u = z^2 + 6 \)
- \( g(u) = -\sin(u) \)
Now, we find \( u' \) and \( g' \):
- \( u' = 2z \)
- \( g' = -\cos(u) \)
Now apply the chain rule:
\[
\frac{d}{dz}(-\sin(z^2 + 6)) = -\cos(z^2 + 6) \cdot 2z
\]
Putting it all together:
\[
g'(z) = 21z^6 - 2z\cos(z^2 + 6)
\]
### Summary of Derivatives
1. \( y' = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2} \)
2. \( h'(\mu) = 10 \sec^2(4 + 10\mu) \)
3. \( G'(x) = 2\cos(3x + \tan(x)) \cdot (3 + \sec^2(x)) \)
4. \( g'(z) = 21z^6 - 2z\cos(z^2 + 6) \)
Quick Answer
1. \( y' = \frac{(3\cos(3t))(1+t^2) - (\sin(3t))(2t)}{(1+t^2)^2} \)
2. \( h'(\mu) = 10 \sec^2(4 + 10\mu) \)
3. \( G'(x) = 2\cos(3x + \tan(x)) \cdot (3 + \sec^2(x)) \)
4. \( g'(z) = 21z^6 - 2z\cos(z^2 + 6) \)
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