Question

Find the derivative of the function. $$ f(x)=6 \sqrt{x}+5 \cos x $$

Short answer

The derivative of the function \(f(x) = 6 \sqrt{x} + 5 \cos x\) is \(f'(x) = 3x^{-1/2} - 5 \sin x\).

Step-by-step solution

Identify and separate terms for differentiation

Recognize that the function \(f(x) = 6 \sqrt{x} + 5 \cos x\) is composed of two distinct terms: \(6 \sqrt{x}\) and \(5 \cos x\). These terms will have to be differentiated separately.

Apply the power rule to differentiate the square root term

The term \(6 \sqrt{x}\) can be rewritten as \(6 x^{1/2}\). Applying the power rule (which states that the derivative of \(x^n\) is \(n*x^{n-1}\)), the derivative of \(6x^{1/2}\) is \(3x^{-1/2}\).

Apply the derivative rule to the cosine term

The derivative of \(5 \cos x\) is \(-5 \sin x\). This comes directly from the knowledge that the derivative of \(\cos x\) is \(-\sin x\).

Combine the derivatives of the terms

Having found the derivatives of the individual terms, combine them to find the derivative of the whole function: \(f'(x) = 3x^{-1/2} - 5 \sin x\)

Key concepts

Power Rule Differentiation

Understanding the power rule is crucial for differentiating a variety of functions in calculus. Simply put, the power rule is used when you encounter a term in the format of \(x^n\), where \(n\) is any real number. The rule states that the derivative of such a term is \(n\) multiplied by the term with its exponent reduced by one. In other words, if you have \(x^n\), then its derivative, denoted as \(\frac{d}{dx}(x^n)\), is \(nx^{n-1}\).

When applying the power rule to a term like \(6x^{1/2}\), which represents \(6\sqrt{x}\), the exponent is \(1/2\). Following the power rule, multiply the exponent by the coefficient to get \(6\times\frac{1}{2} = 3\), and reduce the exponent by one, which is \(1/2 - 1 = -1/2\). Thus, the result is \(3x^{-1/2}\), which signifies the derivative of the square root function in the original exercise. This principle transforms complex expressions into simpler derivatives, enabling us to tackle more intricate calculus problems with ease.

Derivative of Cosine

The derivative of the cosine function is another essential concept in calculus. For any angle \(x\), expressed in radians, the rate of change of the \(\cos x\) function is \(-\sin x\). This means if you are asked to find the derivative of \(\cos x\), you can immediately write down \(-\sin x\) as its derivative without any further calculation.

Understanding Trigonometric Derivatives

Trigonometric functions like sine and cosine have specific derivatives that are essential to know. The fact that the derivative of \(\cos x\) is \(-\sin x\) comes from how the cosine function alters its value as angle \(x\) changes. Recognizing these fundamental trigonometric derivatives is indispensable when solving problems involving trigonometric functions. As illustrated in the exercise, when differentiating \(5\cos x\), you apply the derivative rule to get \(-5\sin x\), where the coefficient \(5\) remains unchanged.

Combining Derivatives

When faced with a function composed of several terms, like \(6\sqrt{x} + 5\cos x\), the strategy to find its derivative is to differentiate each term separately and then combine them. This approach comes from the linearity of the derivative operation, which allows us to take the derivative of terms individually and subsequently add or subtract them according to their original operations.

The power of this method lies in its simplicity and modularity. First, you apply respective differentiation rules to each term: one using the power rule and the other using the rule for differentiating cosine, as covered in the earlier sections. Then, knowing that the original function is simply an addition of these terms, the combined derivative is the addition of their separate derivatives. In our exercise, you differentiate \(6\sqrt{x}\) and \(5\cos x\) separately, obtaining \(3x^{-1/2}\) and \(-5\sin x\) respectively, and combine them. The final derivative of the original function is \(f'(x) = 3x^{-1/2} - 5\sin x\), which succinctly combines the derivatives into a single expression that represents the rate of change of the original function.