Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the derivatives of the following functions. $$f(x)=-\sinh ^{3} 4 x$$

Short Answer

Expert verified
Answer: The derivative of the function $$f(x)=-\sinh ^{3} 4x$$ is $$f'(x)=-12\cosh(4x)\sinh^2(4x)$$.

Step by step solution

01

Break down the function

We can write the given function as follows: $$f(x) = -(\sinh(4x))^3$$ So, we have an outer function $$g(u)=-u^3$$ and an inner function $$u = h(x)=\sinh(4x)$$. Now we need to find the derivatives of both functions with respect to $$x$$.
02

Find the derivative of the outer function

We need to differentiate $$g(u)=-u^3$$ with respect to $$u$$. Using the power rule, we get: $$g'(u)=-3u^2$$
03

Find the derivative of the inner function

We need to differentiate $$h(x)=\sinh(4x)$$ with respect to $$x$$. Using the chain rule, we get: $$h'(x)=\cosh(4x)\cdot 4$$
04

Apply the chain rule

Now, we can use the chain rule to find the derivative of the given function with respect to $$x$$. According to the chain rule, $$f'(x)=g'(h(x))\cdot h'(x)$$ So, plugging in the expressions from Steps 2 and 3, we get: $$f'(x)=-3(\sinh(4x))^2\cdot\cosh(4x)\cdot4$$
05

Simplify the result

Finally, let's simplify the expression: $$f'(x)=-12\cosh(4x)\sinh^2(4x)$$ So, the derivative of the given function is: $$f'(x)=-12\cosh(4x)\sinh^2(4x)$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Chain Rule
The chain rule is a cornerstone of calculus, essential when dealing with composite functions - functions made up of two or more combined functions. Imagine stacking two processes: one follows the other. To find the rate of change for the overall process, the chain rule instructs us to differentiate each function separately and multiply the results.

For a composite function represented as \( f(g(x)) \), the derivative \( f'(x) \) is found by multiplying the derivative of the outer function \( f'(g(x)) \) by the derivative of the inner function \( g'(x) \). This rule allows us to dissect complex expressions and tackle them piece by piece, simplifying the differentiation process.
Hyperbolic Functions
Hyperbolic functions, similar to trigonometric functions, are important in many areas of mathematics including calculus. The hyperbolic sine \( \sinh(x) \) and hyperbolic cosine \( \cosh(x) \) are two fundamental hyperbolic functions.

They're defined using exponential functions as:
\[ \sinh(x) = \frac{e^x - e^{-x}}{2} \] and
\[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
This creates a rich connection to exponential growth and decay processes, making them incredibly useful for modeling real-world phenomena like the shape of a hanging cable or the distribution of heat over time.
Power Rule
The power rule is a straightforward but powerful tool in calculus, used to differentiate functions of the form \( f(x) = x^n \) where \( n \) is any real number. The rule states that the derivative of \( f(x) \) with respect to \( x \) is \( n \) times the function with its power reduced by one:
\[ f'(x) = nx^{n-1} \]
Applying the power rule is like turning down the volume on a function - you decrease its power while amplifying the coefficient. This rule allows rapid computations and is especially handy when dealing with polynomial functions.
Derivative of sinh
Considering \( \sinh(x) \) as a hyperbolic function, its derivative holds a neat relationship with another hyperbolic function. The derivative of \( \sinh(x) \) is \( \cosh(x) \), the hyperbolic cosine of \( x \).

If we have \( \sinh(kx) \) for some constant \( k \), we must remember to multiply by \( k \) due to the chain rule, as seen in \( \sinh(4x) \). So, the derivative of \( \sinh(4x) \) is \( 4 \cosh(4x) \). This captures the essence of calculus - finding patterns and connections which simplify the process of finding rates of change.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the average value of \(f(x)=\frac{1}{x}\) on the interval \([1, p]\) for \(p>1 ?\) What is the average value of \(f\) as \(p \rightarrow \infty ?\)

Designing exponential growth functions Complete the following steps for the given situation. a. Find the rate constant k and use it to devise an exponential growth function that fits the given data. b. Answer the accompanying question. Rising costs Between 2010 and 2016 , the average rate of inflation was about \(1.6 \% / \mathrm{yr}\). If a cart of groceries cost \(\$ 100\) in 2010 , what will it cost in 2025 , assuming the rate of inflation remains constant at \(1.6 \% ?\)

Constant doubling time Prove that the doubling time for an exponentially increasing quantity is constant for all time.

Acceleration, velocity, position Suppose the acceleration of an object moving along a line is given by \(a(t)=-k v(t),\) where \(k\) is a positive constant and \(v\) is the object's velocity. Assume the initial velocity and position are given by \(v(0)=10\) and \(s(0)=0\) respectively. a. Use \(a(t)=v^{\prime}(t)\) to find the velocity of the object as a function of time. L. Use \(v(t)=s^{\prime}(t)\) to find the position of the object as a function of time. c. Use the fact that \(\frac{d v}{d t}=\frac{d v}{d s} \frac{d s}{d t}\) (by the Chain Rule) to find the velocity as a function of position.

Evaluate the following definite integrals. Use Theorem 7.7 to express your answer in terms of logarithms. $$\int_{\ln 5}^{\ln 9} \frac{\cosh x}{4-\sinh ^{2} x} d x$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free