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To determine the expression for the derivative dy/dx of the function y = (4x + 1)ln(3x + 1), one must apply the product rule of differentiation, as the function is a product of two distinct components: (4x + 1) and ln(3x + 1). The product rule states that if you have two functions u and v, the derivative of their product is given by u’v + uv’. In this case, let u = (4x + 1) and v = ln(3x + 1). The derivative of u with respect to x, denoted as u’, is 4, while the derivative of v, denoted as v’, requires the chain rule, yielding v’ = (1/(3x + 1))(3) = 3/(3x + 1). By substituting these derivatives into the product rule formula, one can derive the expression for dy/dx.
After applying the product rule, the expression for dy/dx can be simplified. This results in dy/dx = u’v + uv’ = 4ln(3x + 1) + (4x + 1)(3/(3x + 1)). Further simplification of the second term leads to dy/dx = 4ln(3x + 1) + (12x + 3)/(3x + 1). This expression encapsulates the rate of change of the function y with respect to x, providing a comprehensive understanding of its behaviour across different values of x.
To find the gradient of the function at the specific point where x = 1, one must substitute x = 1 into the derived expression for dy/dx. By performing this substitution, one calculates dy/dx at x = 1, which yields the gradient at that point. This process not only provides the numerical value of the gradient but also illustrates the instantaneous rate of change of the function y at x = 1, thereby offering insights into the function’s local behaviour at that particular coordinate.
