Question Statement

Find the derivative of the function $h(t)=\frac{t+\sin 4t}{10+\cos 3t}$ with respect to $t$.

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Background and Explanation

This problem requires applying the quotient rule for differentiation, combined with the chain rule to handle the composite trigonometric functions $\sin 4t$ and $\cos 3t$.

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Solution

Given the function: $h(t)=\frac{t+\sin 4t}{10+\cos 3t}$

Differentiate with respect to $t$ using the quotient rule. Let $u=t+\sin 4t$ and $v=10+\cos 3t$.

$\begin{array}{rl}{h}^{\prime }(t)& =\frac{(10+\cos 3t)\frac{d}{dt}(t+\sin 4t)-(t+\sin 4t)\frac{d}{dt}(10+\cos 3t)}{(10+\cos 3t{)}^2}\\ & =\frac{(10+\cos 3t)(1+\cos 4t\cdot 4)-(t+\sin 4t)\cdot (0+(-\sin 3t)\cdot 3)}{(10+\cos 3t{)}^2}\\ & =\frac{(10+\cos 3t)(1+4\cos 4t)-(t+\sin 4t)(3\sin 3t)}{(10+\cos 3t{)}^2}\end{array}$

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dt}\left[\frac{u(t)}{v(t)}\right]=\frac{v(t)\cdot u^{\prime }(t)-u(t)\cdot v^{\prime }(t)}{[v(t){]}^2}$
• Chain Rule: $\frac{d}{dt}[f(g(t))]=f^{\prime }(g(t))\cdot g^{\prime }(t)$
• Derivative of Sine: $\frac{d}{dt}(\sin kt)=k\cos kt$
• Derivative of Cosine: $\frac{d}{dt}(\cos kt)=-k\sin kt$

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Summary of Steps

1. Identify components: Let $u=t+\sin 4t$ (numerator) and $v=10+\cos 3t$ (denominator).
2. Differentiate the numerator: $\frac{d}{dt}(t+\sin 4t)=1+4\cos 4t$ using the chain rule.
3. Differentiate the denominator: $\frac{d}{dt}(10+\cos 3t)=-3\sin 3t$ using the chain rule.
4. Apply the quotient rule: Substitute $u,v,u^{\prime },$ and $v^{\prime }$ into the formula $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$.
5. Simplify the expression: Expand the terms in the numerator and write the final derivative.