Let's solve the derivatives step by step for each part of the given problems:

a) $y = 5\sqrt[5]{x^4} + 10^x - 7$

• Rewrite $5\sqrt[5]{x^4}$ as $5x^{4/5}$.

The derivative is: $y' = 5 \cdot \frac{4}{5} x^{4/5 - 1} + 10^x \ln(10) + 0$

Simplify: $y' = 4x^{-1/5} + 10^x \ln(10)$

b) $y = \frac{1 + 3^x}{1 - 3^x}$

Use the quotient rule: $y' = \frac{(1 - 3^x) \cdot \frac{d}{dx}(1 + 3^x) - (1 + 3^x) \cdot \frac{d}{dx}(1 - 3^x)}{(1 - 3^x)^2}$

Compute the derivatives in the numerator: $y' = \frac{(1 - 3^x) \cdot (0 + 3^x \ln(3)) - (1 + 3^x) \cdot (0 - 3^x \ln(3))}{(1 - 3^x)^2}$

Simplify: $y' = \frac{3^x \ln(3)(1 - 3^x) + 3^x \ln(3)(1 + 3^x)}{(1 - 3^x)^2}$

Combine terms: $y' = \frac{3^x \ln(3) \cdot 2}{(1 - 3^x)^2}$

c) $y = \cos\left(1 + \sqrt{x}\right)$

Use the chain rule: $y' = -\sin\left(1 + \sqrt{x}\right) \cdot \frac{d}{dx}\left(1 + \sqrt{x}\right)$

Compute the inner derivative: $y' = -\sin\left(1 + \sqrt{x}\right) \cdot \frac{1}{2\sqrt{x}}$

Simplify: $y' = -\frac{\sin\left(1 + \sqrt{x}\right)}{2\sqrt{x}}$

d) Parametric equations:

$x = t^4 - 4t, \quad y = \arctan\left(\frac{1}{t}\right)$

We use the chain rule for $y'_x = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

1. Compute $\frac{dx}{dt}$: $\frac{dx}{dt} = 4t^3 - 4$

2. Compute $\frac{dy}{dt}$: $y = \arctan\left(\frac{1}{t}\right) \quad \implies \quad \frac{dy}{dt} = \frac{1}{1 + \left(\frac{1}{t}\right)^2} \cdot \left(-\frac{1}{t^2}\right)$

Simplify $\frac{dy}{dt}$: $\frac{dy}{dt} = \frac{-1}{t^2 + 1}$

3. Compute $y'_x$: $y'_x = \frac{\frac{-1}{t^2 + 1}}{4t^3 - 4}$

Simplify: $y'_x = \frac{-1}{(t^2 + 1)(4t^3 - 4)}$

Do you need further explanations or more steps detailed?

Related Questions:

1. What is the chain rule, and how is it applied in derivative calculations?
2. How do you simplify derivatives involving fractional exponents, such as in problem (a)?
3. Why does the logarithmic function appear in derivatives of exponential terms like $10^x$?
4. How do you approach derivatives for parametric equations, like in problem (d)?
5. What techniques are essential for solving trigonometric derivatives, such as $\cos$ and $\arctan$?

Tip:

When solving derivatives, always simplify terms step by step before applying the chain or quotient rule. It reduces errors and makes the computation clearer.