Let’s differentiate each expression using the chain rule.

Part (a): $f(x) = (3x^2 - 5x + 2)^4$

1. Let $u = 3x^2 - 5x + 2$, so $f(x) = u^4$.
2. Differentiate $f(x) = u^4$ with respect to $u$: $\frac{df}{du} = 4u^3$
3. Now, differentiate $u = 3x^2 - 5x + 2$ with respect to $x$: $\frac{du}{dx} = 6x - 5$
4. Apply the chain rule: $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = 4u^3 \cdot (6x - 5)$
5. Substitute back $u = 3x^2 - 5x + 2$: $\frac{df}{dx} = 4(3x^2 - 5x + 2)^3 \cdot (6x - 5)$

Part (b): $g(x) = \frac{1}{3x} \cdot 7$

1. Rewrite $g(x) = \frac{7}{3x}$, which simplifies to $g(x) = \frac{7}{3} \cdot x^{-1}$.
2. Differentiate with respect to $x$: $g'(x) = \frac{7}{3} \cdot (-1) \cdot x^{-2} = -\frac{7}{3x^2}$

Part (c): $h(x) = \sqrt{1 + x^2}$

1. Rewrite $h(x) = (1 + x^2)^{1/2}$.
2. Let $u = 1 + x^2$, so $h(x) = u^{1/2}$.
3. Differentiate $h(x) = u^{1/2}$ with respect to $u$: $\frac{dh}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$
4. Now, differentiate $u = 1 + x^2$ with respect to $x$: $\frac{du}{dx} = 2x$
5. Apply the chain rule: $\frac{dh}{dx} = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{u}} \cdot 2x$
6. Substitute back $u = 1 + x^2$: $\frac{dh}{dx} = \frac{x}{\sqrt{1 + x^2}}$

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Related Questions

1. How can the product rule and chain rule be combined in differentiation?
2. What’s the derivative of $\sin^2(x)$ using the chain rule?
3. How does the chain rule apply in multivariable calculus?
4. Can we use the chain rule for implicit differentiation?
5. What are some common mistakes when applying the chain rule?

Tip

When differentiating compositions of functions, always start by identifying the "outer" and "inner" functions to apply the chain rule effectively.