Problem 64 - Single Variable Calculus

Question

Find the derivative of the function. $$ g(x)=\frac{2}{\sqrt{3^{x}+1}} $$

Step-by-Step Solution

Answer

The derivative of \(g(x) = \frac{2}{\sqrt{3^x+1}}\) is \(g'(x) = -\frac{3^x \ln{3}}{2(3^x + 1)^{\frac{3}{2}}}\).

Step 1: Rewrite the function in a suitable form for differentiation

Express the square root in terms of an exponent and rewrite the function: $$ g(x) = (3^x + 1)^{-\frac{1}{2}} $$ This will make differentiation easier.

Step 2: Apply the Chain Rule

Now, we can differentiate g(x) with respect to x using the Chain Rule. Let's denote the inner function as u: $$ u(x) = 3^x + 1 $$ We will first find the derivative of the outer function with respect to u, which involves differentiating \((u)^{-\frac{1}{2}}\) w.r.t u and then substituting x back later: $$ \frac{d}{du} (u)^{-\frac{1}{2}} = -\frac{1}{2}(u)^{-\frac{3}{2}} $$ Now we need to find the derivative of the inner function with respect to x: $$ \frac{du}{dx} = \frac{d}{dx} (3^x + 1) = 3^x \ln{3} $$ Now we can find the derivative of g(x) by applying the Chain Rule: $$ \frac{dg}{dx} = \frac{d}{du} (u)^{-\frac{1}{2}} \times \frac{du}{dx} $$ Substitute the values of the derivatives we found before:

Step 3: Calculate the derivative and simplify

Now we can substitute the inner function back and find the derivative: $$ \frac{dg}{dx} = -\frac{1}{2}(3^x + 1)^{-\frac{3}{2}} \times 3^x \ln{3} $$ Finally, we can simplify the derivative expression: $$ g'(x) = -\frac{3^x \ln{3}}{2(3^x + 1)^{\frac{3}{2}}} $$ So, the derivative of the given function is: $$ g'(x) = -\frac{3^x \ln{3}}{2(3^x + 1)^{\frac{3}{2}}} $$

Key Concepts

Chain Rule

The Chain Rule is a powerful tool in calculus for finding the derivative of composite functions. Imagine you have a function within another function, like peeling layers of an onion. To differentiate a composite function, we apply this rule by working with each layer separately.

For the function \(g(x) = (3^x + 1)^{-\frac{1}{2}}\), we first identify the inner function \(u(x) = 3^x + 1\). The outer function is \(v(u) = u^{-\frac{1}{2}}\).

• Differentiate the outer function with respect to the inner function \(u\).
• Differentiate the inner function \(u\) with respect to \(x\).
• Multiply these two derivatives together to get the final result.

Using the Chain Rule simplifies this complex process by breaking it into manageable pieces.

Derivative of Exponential Functions

Exponential functions, like \(3^x\), have unique properties that require special attention when differentiating. These functions grow rapidly, and their derivatives are directly tied to the function itself.

The derivative of an exponential function \(a^x\) is \(a^x \ln{a}\). For \(g(x) = (3^x + 1)^{-\frac{1}{2}}\), we need to find \(\frac{d}{dx}(3^x)\), which is \(3^x \ln{3}\).

• The constant \(\ln{a}\) comes from shifting the exponential base.
• Only apply this rule when the base is the same as in the function.

Understanding these nuances helps you handle any exponential derivative you encounter.

Derivative Simplification

Simplifying your derivative is the final step to making your answer clear and concise. In our example, simplification involves rewriting the found derivative in a more elegant form.

After applying the Chain Rule, we found the derivative to be \(\frac{dg}{dx} = -\frac{1}{2}(3^x + 1)^{-\frac{3}{2}} \times 3^x \ln{3}\). By substituting \(g'(x)\) back into a simplified structure, we arrive at:
\[g'(x) = -\frac{3^x \ln{3}}{2(3^x + 1)^{\frac{3}{2}}}\]

• Combine constants and terms to make the expression cleaner.
• Factor out common elements if possible.

Simplification not only makes your answer neat but also facilitates easier interpretation and further computations.