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Use a CAS to find \(d y / d x\). \(y=\left[x \sin 2 x+\tan ^{4}\left(x^{7}\right)\right]^{5}\)

Short Answer

Expert verified
The derivative is \( \frac{dy}{dx} = 5[x \sin 2x + \tan^4(x^7)]^4 \cdot (\sin 2x + 2x \cos 2x + 28x^6 \tan^3(x^7) \sec^2(x^7)). \)

Step by step solution

01

Identify the Outer Function

The given function is \( y = [x \sin 2x + \tan^4(x^7)]^5 \). Here, the outermost function is something raised to the power of 5.
02

Apply the Chain Rule

To differentiate the function \([u]^5\), we use the chain rule: \( \frac{dy}{dx} = 5[u]^4 \cdot \frac{du}{dx} \). Let \( u = x \sin 2x + \tan^4(x^7) \).
03

Differentiate the Inner Function

Differentiate \( u = x \sin 2x + \tan^4(x^7) \) with respect to \( x \):\[ \frac{du}{dx} = \frac{d}{dx}[x \sin 2x] + \frac{d}{dx}[\tan^4(x^7)]. \]
04

Differentiate Each Term

- For \( \frac{d}{dx}[x \sin 2x] \), use the product rule: \( \frac{d}{dx}[x] \cdot \sin 2x + x \cdot \frac{d}{dx}[\sin 2x] = \sin 2x + 2x \cos 2x \).- For \( \frac{d}{dx}[\tan^4(x^7)] \), apply the chain and power rules: \( 4 \tan^3(x^7) \cdot \sec^2(x^7) \cdot 7x^6 \).
05

Combine the Results

Combine the derivatives to get \( \frac{du}{dx} = \sin 2x + 2x \cos 2x + 28x^6 \tan^3(x^7) \sec^2(x^7) \).
06

Substitute Back into Chain Rule

Substitute \( \frac{du}{dx} \) back into the chain rule formula: \[ \frac{dy}{dx} = 5[x \sin 2x + \tan^4(x^7)]^4 \cdot \left(\sin 2x + 2x \cos 2x + 28x^6 \tan^3(x^7) \sec^2(x^7)\right). \]

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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 concept in calculus that allows us to differentiate composite functions. A composite function is essentially a function within another function. In simpler terms, if you have a function defined as \(y = f(g(x))\), the chain rule helps differentiate this by focusing on two things:
  • The derivative of the outer function, \(f\), evaluated at the inner function, \(g(x)\).
  • The derivative of the inner function, \(g(x)\), with respect to \(x\).
The chain rule formula can be expressed as \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \].
Using this rule allows us to find derivatives of complex functions step by step. In the given exercise, the general framework is to first identify the outer function and treat the inner part as a single unit. Then, find the derivative of that inner part separately. Finally, combine them using multiplication to find the derivative of the entire function.
Product Rule
When dealing with functions that are multiplied together, the product rule is essential. This rule states how to differentiate products of two or more functions. For two functions \(u(x)\) and \(v(x)\), the product rule is given by:
  • The first function's derivative multiplied by the second function unchanged.
  • Plus the first function unchanged multiplied by the second function's derivative.
Mathematically, this is described as \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \].
In our specific exercise, we apply the product rule to differentiate the term \(x \sin 2x\).
Here, \(x\) is the first function and \(\sin 2x\) is the second.
Differentiating each part separately, you then add them according to the product rule to find the derivative of the whole product part effectively.
Trigonometric Differentiation
Trigonometric differentiation involves finding derivatives of trigonometric functions. These functions include \(\sin\), \(\cos\), \(\tan\), and their derivatives are essential for solving many calculus problems.
For basic trigonometric functions, the derivatives are as follows:
  • The derivative of \(\sin x\) is \(\cos x\).
  • The derivative of \(\cos x\) is \(-\sin x\).
  • The derivative of \(\tan x\) is \(\sec^2 x\).
In the problem, we encounter \(\tan^4(x^7)\) where advanced differentiation techniques such as the power rule and chain rule were incorporated. The derivative \(\frac{d}{dx}[\tan^4(x^7)]\) involves understanding that \(\tan^4(x^7)\) is \((\tan(x^7))^4\).
Here, by applying a combination of chain and power rules, the differentiation is simplified into manageable parts, yielding a clear and concise derivative function.

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Most popular questions from this chapter

Find the \(x\) -coordinate of the point on the graph of \(y=x^{2}\) where the tangent line is parallel to the secant line that cuts the curve at \(x=-1\) and \(x=2\).

Find an equation for the tangent line to the graph at the specified value of \(x\). \(y=\sin \left(1+x^{3}\right), x=-3\)

(a) Use a graphing utility to obtain the graph of the function \(f(x)=\sin x^{2} \cos x\) over the interval \([-\pi / 2, \pi / 2]\) (b) Use the graph in part (a) to make a rough sketch of the graph of \(f^{\prime}\) over the interval. (c) Find \(f^{\prime}(x)\), and then check your work in part (b) by using the graphing utility to obtain the graph of \(f^{\prime}\) over the interval. (d) Find the equation of the tangent line to the graph of \(f\) at \(x=1\), and graph \(f\) and the tangent line together over the interval.

The "co" in "cosine" comes from "complementary," since the cosine of an angle is the sine of the complementary angle, and vice versa: $$ \cos x=\sin \left(\frac{\pi}{2}-x\right) \text { and } \sin x=\cos \left(\frac{\pi}{2}-x\right) $$ Suppose that we define a function \(g\) to be a cofunction of a function \(f\) if $$ g(x)=f\left(\frac{\pi}{2}-x\right) \quad \text { for all } x $$ Thus, cosine and sine are cofunctions of each other, as are cotangent and tangent, and also cosecant and secant. If \(g\) is the cofunction of \(f\), state a formula that relates \(g^{\prime}\) and the cofunction of \(f^{\prime}\). Discuss how this relationship is exhibited by the derivatives of the cosine, cotangent, and cosecant functions.

Find an equation for the tangent line to the graph at the specified value of \(x\). \(y=\frac{x}{\sqrt{1-x^{2}}}, x=0\)

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