When differentiating products of two functions, the Product Rule must be used.

Composite functions are differentiated using the Chain Rule. To do this, the outside function and the inside function must be identified. The derivative of the composite function is the product of the derivative of the outside function and the derivative of the inside function.

Fractions of functions are differentiated using the Quotient rule.

To find the derivative of the inverse of a function, the inverse of the function and the derivative of it must be found. Then, the inverse must be substituted into the derivative of the function. Taking the reciprocal of this result gives the derivative of the inverse of the function. This is called the Inverse Function Theorem.

The given function can be differentiated using chain rule.
Let y = u⁵ and u = 1 + x⁴.
Then, dy/du = 5u⁴ and du/dx = 4x³.
By the chain rule, dy/dx = 5u⁴ × 4x³ = 20x³(1 + x⁴)⁴.

20x³(1 + x⁴)⁴

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f'(x) = 35x⁶ and f⁻¹(x) = ⁷√(x/5). So, d(f⁻¹(x))/dx = 1/(35(⁷√(x/5))⁶) = 1/(35(x/5)⁶/⁷).

The question can be solved by using the quotient rule.
Let u = (5x² + 12x)² and v = 2x + 1.
Then, by the chain rule, du/dx = 2(5x² + 12x)(10x + 12) and dv/dx = 2.
By the quotient rule, dy/dx = (2(5x² + 12x)(10x + 12)(2x + 1) − 2(5x² + 12x)²)/(2x + 1)².
dy/dx = 2x(75x³ + 290x² + 324x + 144)/(2x + 1)²

2x(75x³ + 290x² + 324x + 144)/(2x + 1)²

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Area = t³(t + 1) and Perimeter = t³ + t³ + t + 1 + t + 1 = 2t³ + 2t + 2. So, we want to differentiate f(t) = t³(t + 1)/(2t³ + 2t + 2). This can be done using the Quotient rule. Let u = t³(t + 1) and v = 2t³ + 2t + 2. Then, by the Product rule, du/dt = t³ + 3t²(t + 1) = 4t³ + 3t² and dv/dt = 6t² + 2. So, by the Quotient rule, f'(t) = ((4t³ + 3t²)(2t³ + 2t + 2) − t³(t + 1)(6t² + 2))/(2t³ + 2t + 2)².

Let y = u² and u = x² + 8x + 27.
Then, dy/du = 2u and du/dx = 2x + 8.
By the chain rule, dy/dx = 2u × (2x + 8) = 2(2x + 8)(x² + 8x + 27).

2(2x + 8)(x² + 8x + 27)