Differentiating Trigonometric, Logarithmic, and Exponential Functions

To successfully solve differentiation problems involving trigonometric, logarithmic, and exponential functions, a student must begin by memorizing their core derivative formulas. For trigonometric functions, the fundamental derivatives are `\frac{d}{dx}(\sin(x)) = \cos(x)` and `\frac{d}{dx}(\cos(x)) = -\sin(x)`. For exponential and logarithmic functions, the natural forms are the easiest to work with: the derivative of `e^x` remains `e^x`, while the derivative of the natural logarithm `\ln(x)` is `\frac{1}{x}`. If the base is a constant other than `e`, such as `a^x` or `\log_a(x)`, you must adjust the derivative by multiplying or dividing by `\ln(a)` respectively.

The true challenge arises when these transcendental functions are nested within other expressions, which requires a rigorous application of the Chain Rule, Product Rule, or Quotient Rule. The Chain Rule, expressed as `\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)`, dictates that you must differentiate the outer "shell" function while keeping the inner function intact, and then multiply by the derivative of that inner function. When trigonometric or exponential terms are multiplied by polynomials or each other, the Product Rule `(uv)' = u'v + uv'` ensures that every part of the expression is correctly accounted for during the differentiation process.

Let us look at a practical example by finding the derivative of the function `f(x) = e^{2x} \cdot \ln(\sin(x))`. This problem requires a combination of the Product Rule and the Chain Rule. First, applying the Product Rule, we take the derivative of the first term, which by the Chain Rule is `2e^{2x}`, and multiply it by the second term to get `2e^{2x} \cdot \ln(\sin(x))`. Next, we keep the first term `e^{2x}` and multiply it by the derivative of `\ln(\sin(x))`. The derivative of `\ln(\sin(x))` requires the Chain Rule: the outer derivative is `\frac{1}{\sin(x)}` and the inner derivative of `\sin(x)` is `\cos(x)`, giving `\frac{\cos(x)}{\sin(x)} = \cot(x)`. Combining these parts, the final derivative is `f'(x) = 2e^{2x}\ln(\sin(x)) + e^{2x}\cot(x)`.