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Chapter 3: Problem 49

Calculate the derivative of the following functions. $$\tan \left(e^{\sqrt{3 x}}\right)$$

Short Answer

Expert verified
Question: Find the derivative of the function $$\tan \left(e^{\sqrt{3 x}}\right)$$ with respect to x. Answer: The derivative of the given function is $$\frac{d}{dx}\tan\left(e^{\sqrt{3 x}}\right) = \frac{3\sec^2\left(e^{\sqrt{3 x}}\right)e^{\sqrt{3 x}}}{2\sqrt{3x}}$$.

Step by step solution

01

Identify the outer function

The outer function is: $$f(u) = \tan(u)$$ Let $$u = e^{\sqrt{3 x}}$$
02

Differentiate the outer function

The derivative of the outer function with respect to u is: $$f'(u) = \frac{d}{du}\tan(u) = \sec^2(u)$$
03

Identify the inner function

The inner function is: $$u = g(x) = e^{\sqrt{3 x}}$$
04

Differentiate the inner function

Applying chain rule, the derivative of the inner function with respect to x is: $$g'(x) = \frac{d}{dx} e^{\sqrt{3 x}} = e^{\sqrt{3 x}} \cdot \frac{d}{dx}[\sqrt{3 x}]$$ Now, we need to differentiate $$\sqrt{3 x}$$. Using the power rule, we have: $$\frac{d}{dx}[\sqrt{3 x}] = \frac{1}{2}(3x)^{-\frac{1}{2}}\cdot 3 = \frac{3}{2(3x)^{\frac{1}{2}}} = \frac{3}{2\sqrt{3x}}$$ So, we have: $$g'(x) = e^{\sqrt{3 x}} \cdot \frac{3}{2\sqrt{3x}}$$
05

Apply the chain rule

Now that we have the derivatives of the outer and inner functions, we can apply the chain rule: $$\frac{d}{dx}\tan\left(e^{\sqrt{3 x}}\right) = f'(g(x))\cdot g'(x)$$ Substitute the values we've obtained: $$\frac{d}{dx}\tan\left(e^{\sqrt{3 x}}\right) = \sec^2\left(e^{\sqrt{3 x}}\right)\cdot\left[e^{\sqrt{3 x}} \cdot \frac{3}{2\sqrt{3x}}\right]$$
06

Simplify the expression

Now, just simplify the final expression: $$\frac{d}{dx}\tan\left(e^{\sqrt{3 x}}\right) = \frac{3\sec^2\left(e^{\sqrt{3 x}}\right)e^{\sqrt{3 x}}}{2\sqrt{3x}}$$ So the derivative of the given function is: $$\frac{d}{dx}\tan\left(e^{\sqrt{3 x}}\right) = \frac{3\sec^2\left(e^{\sqrt{3 x}}\right)e^{\sqrt{3 x}}}{2\sqrt{3x}}$$

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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 fundamental theorem in calculus that allows us to differentiate composite functions.
This means if you have a function nested inside another function, chain rule helps in finding the derivative.
  • When you look at a function like \( \tan(e^{\sqrt{3x}}) \), you can see there鈥檚 an outer function \( \tan(u) \) and an inner function \( e^{\sqrt{3x}} \).
  • The process involves differentiating the outer function with respect to the inner function first.
  • Then, you differentiate the inner function with respect to the main variable, often \( x \).
Using the chain rule involves setting up this derivative by multiplying these two results.
It is a key component for understanding more complex differentiation techniques in calculus.
Outer and Inner Functions
Identifying outer and inner functions is crucial for applying the chain rule.
  • The outer function is the function on the outside of your composite function, here it鈥檚 \( \tan(u) \).
  • The inner function is the one inside, in this case, \( e^{\sqrt{3x}} \).
Once identified, differentiate each separately.
  • Differentiate the outer function with respect to the inner: \( \dfrac{d}{du} \tan(u) = \sec^2(u) \).
  • For the inner function, since it is exponential, use rules for derivatives like exponential and power rule combined.
Getting comfortable with identifying these functions helps a lot in mastering the chain rule.
It's like peeling an onion, taking the outer layer (function) off first, then tackling what's inside.
Simplifying Expressions
After applying the chain rule and determining the derivatives, we're left with an expression that needs simplifying.
  • The derivative \( \sec^2(e^{\sqrt{3x}}) \cdot \left[e^{\sqrt{3x}} \cdot \dfrac{3}{2\sqrt{3x}}\right] \) combines products and fractions.
  • Simplification involves carefully multiplying through these components.
The goal is to express the derivative as neatly as possible.
  • Ensure all similar terms are reduced or combined, like constants and similar bases.
  • Simplifying doesn鈥檛 just make the result pretty, it's crucial for interpreting and further application.
Here, the process gives the final simplified derivative: \( \dfrac{3 \sec^2\left(e^{\sqrt{3 x}}\right) e^{\sqrt{3 x}}}{2\sqrt{3x}} \).
Learning this part ensures that we communicate our findings effectively and use them in wider mathematical contexts.

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Most popular questions from this chapter

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}(f(x) g(x))\right|_{x=1}$$

Means and tangents Suppose \(f\) is differentiable on an interval containing \(a\) and \(b,\) and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x},\) show that \(c=\sqrt{a b},\) the geometric mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{x}{x+2}\)

Find the slope of the curve \(5 \sqrt{x}-10 \sqrt{y}=\sin x\) at the point \((4 \pi, \pi)\).

An angler hooks a trout and begins turning her circular reel at \(1.5 \mathrm{rev} / \mathrm{s}\). If the radius of the reel (and the fishing line on it) is 2 in. then how fast is she reeling in her fishing line?
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