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Chapter 2: Problem 64

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

Short Answer

Expert verified
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 by step solution

01

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.
02

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:
03

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: $$ g'(x) = -\frac{3^x \ln{3}}{2(3^x + 1)^{\frac{3}{2}}} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Forecasting Commodity Crops Government economists in a certain country have determined that the demand equation for soybeans is given by $$ p=f(x)=\frac{55}{2 x^{2}+1} $$ where the unit price \(p\) is expressed in dollars per bushel and \(x\), the quantity demanded per year, is measured in billions of bushels. The economists are forecasting a harvest of \(2.2\) billion bushels for the year, with a possible error of \(10 \%\) in their forecast. Determine the corresponding error in the predicted price per bushel of soybeans.

Electric Potential Suppose that a ring-shaped conductor of radius \(a\) carries a total charge \(Q\). Then the electrical potential at the point \(P\), a distance \(x\) from the center and along the line perpendicular to the plane of the ring through its center, is given by $$ V(x)=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\sqrt{x^{2}+a^{2}}} $$ where \(\varepsilon_{0}\) is a constant called the permittivity of free space. The magnitude of the electric field induced by the charge at the point \(P\) is \(E=-d V / d x\), and the direction of the field is along the \(x\) -axis. Find \(E\).

Defermine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(f\) is differentiable and \(h(t)=f(a+b t)+f(a-b t)\), then \(h^{\prime}(t)=b f^{\prime}(a+b t)-b f^{\prime}(a-b t) .\)

Velocity of a Ballast A ballast of mass \(m\) slugs is dropped from a hot-air balloon with an initial velocity of \(v_{0} \mathrm{ft} / \mathrm{sec}\). If the ballast is subjected to air resistance that is directly proportional to its instantaneous velocity, then its velocity at time \(t\) is $$ v(t)=\frac{m g}{k}+\left(v_{0}-\frac{m g}{k}\right) e^{-k t / m} $$ feet per second, where \(k>0\) is the constant of proportionality and \(g\) is the constant of acceleration. Find an expression for the velocity of the ballast at any time \(t\), assuming that there is no air resistance. Hint: Find \(\lim _{k \rightarrow 0} v(t)\).

Find the derivative of the function. $$ f(x)=\sin ^{-1} 3 x $$
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