Step-by-Step Guidance for Derivatives of Composite Functions

Q6(a). Find the derivative of $r(x) = \sin(\cos(x))$

Background

Topic: Chain Rule for Derivatives

This question tests your ability to differentiate composite functions using the chain rule. Here, $\sin(\cos(x))$ is a composition of the sine and cosine functions.

Key Terms and Formulas

• Chain Rule: If $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$

• $\frac{d}{dx}[\sin(u)] = \cos(u) \cdot u'$

• $\frac{d}{dx}[\cos(x)] = -\sin(x)$

Step-by-Step Guidance

1. Identify the outer function $f(u) = \sin(u)$ and the inner function $g(x) = \cos(x)$.

2. Apply the chain rule: $r'(x) = f'(g(x)) \cdot g'(x)$.

3. Compute $f'(u)$: The derivative of $\sin(u)$ is $\cos(u)$.

4. Compute $g'(x)$: The derivative of $\cos(x)$ is $-\sin(x)$.

5. Set up the expression for $r'(x)$ using these derivatives, but do not simplify to the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$r'(x) = \cos(\cos(x)) \cdot (-\sin(x)) = -\sin(x) \cos(\cos(x))$

This result comes from applying the chain rule to the composite function.

Q6(b). Find the derivative of $h(x) = \tan(2x^2 - 4x + 3)$

Background

Topic: Chain Rule and Derivative of Tangent Function

This question tests your ability to differentiate a tangent function whose argument is a quadratic expression, requiring the chain rule.

Key Terms and Formulas

• Chain Rule: If $y = f(g(x))$, then $y' = f'(g(x)) \cdot g'(x)$

• $\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot u'$

• $\frac{d}{dx}[2x^2 - 4x + 3] = 4x - 4$

Step-by-Step Guidance

1. Identify the outer function $f(u) = \tan(u)$ and the inner function $g(x) = 2x^2 - 4x + 3$.

2. Apply the chain rule: $h'(x) = f'(g(x)) \cdot g'(x)$.

3. Compute $f'(u)$: The derivative of $\tan(u)$ is $\sec^2(u)$.

4. Compute $g'(x)$: The derivative of $2x^2 - 4x + 3$ is $4x - 4$.

5. Set up the expression for $h'(x)$ using these derivatives, but do not simplify to the final answer yet.

Try solving on your own before revealing the answer!

Final Answer:

$h'(x) = \sec^2(2x^2 - 4x + 3) \cdot (4x - 4)$

This result is obtained by applying the chain rule to the tangent function with a quadratic argument.