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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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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