/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 Proof of \(\frac{d}{d x}(\cos x)... [FREE SOLUTION] | 91Ó°ÊÓ

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Proof of \(\frac{d}{d x}(\cos x)=-\sin x\) Use the definition of the derivative and the trigonometric identity $$ \cos (x+h)=\cos x \cos h-\sin x \sin h $$ to prove that \(\frac{d}{d x}(\cos x)=-\sin x\)

Short Answer

Expert verified
The derivative of the cosine function is equal to the negative sine function, represented as \(\frac{d}{d x}(\cos x) = -\sin x\).

Step by step solution

01

Write the definition of the derivative for cos(x)

We will use the definition of the derivative for the cosine function by finding the limit: $$ \frac{d}{d x}(\cos x) = \lim_{h \to 0} \frac{\cos (x+h)-\cos x}{h} $$
02

Use the trigonometric identity to simplify the expression

Now, we will substitute the given trigonometric identity for \(\cos (x+h)\) in the expression: $$ \frac{d}{d x}(\cos x) = \lim_{h \to 0} \frac{(\cos x \cos h - \sin x \sin h) - \cos x}{h} $$
03

Simplify and separate the expressions for sin(x) and cos(x)

Simplify the expression above and separate the parts containing \(\sin x\) and \(\cos x\): $$ \frac{d}{d x}(\cos x) = \lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x (\sin h)}{h} $$
04

Split the limit into two separate limits

Now, we will split the expression into two separate limits: $$ \frac{d}{d x}(\cos x) = \cos x \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \lim_{h \to 0} \frac{\sin h}{h} $$
05

Evaluate the limits

Using the well-known limits \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\) and \(\lim_{h \to 0} \frac{\sin h}{h} = 1\), we have: $$ \frac{d}{d x}(\cos x) = \cos x (0) - \sin x (1) $$
06

Reach the final result

Simplify the expression above to obtain the derivative of the cosine function: $$ \frac{d}{d x}(\cos x) = -\sin x $$ This proves that \(\frac{d}{d x}(\cos x) = -\sin x\).

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Key Concepts

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

Definition of the Derivative
The derivative of a function provides us with the rate at which the function changes at any given point. In simple terms, it tells us how steep the curve of the function is at a particular spot.
To compute the derivative of a function at a point, we use a formula involving limits. This formula is known as the definition of the derivative. It is expressed as:
\(\frac{d}{dx}f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
In this formula, \(h\) symbolizes a small change in the \(x\) value. If we think about \(x+h\), it means we move slightly away from \(x\).
Taking the limit as \(h\) approaches zero means looking at what happens as the change in \(x\) gets smaller and smaller. This way, we find the instantaneous rate of change at \(x\). For our specific example, we are dealing with \(\cos(x)\), and our goal is to find how \(\cos(x)\) changes as \(x\) changes.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are always true for any angle. These identities help simplify expressions and solve trigonometric equations.
One of the most useful identities in calculus is the angle addition formula for cosine:
\(\cos(x+h) = \cos x \cos h - \sin x \sin h\).
This identity lets us break down complex trigonometric expressions into simpler parts that are easier to work with.
In the context of derivatives, using these identities allows us to rearrange and manipulate functions to apply the limit definition effectively.
By substituting the identity into the derivative of \(\cos(x)\), we can easily see how the pieces fit together and apply limits to find the derivative.
Limit of a Function
The limit of a function describes what happens when we get closer and closer to a particular point on the function. In calculus, limits are foundational because they help us understand behavior near specific points.
Limits are expressed as:
\(\lim_{h \to 0} f(h)\)
This notation means we are interested in what value \(f(h)\) approaches as \(h\) becomes incredibly small.
For instance, in the context of derivatives, we often come across limits like \(\lim_{h \to 0} \frac{\cos h - 1}{h} = 0\) and \(\lim_{h \to 0} \frac{\sin h}{h} = 1\).
These specific limits are crucial in finding the derivative of trigonometric functions, letting us simplify expressions by knowing their behavior as \(h\) approaches zero.
Through limits, we gain a precise mathematical way to describe rates of change and instantaneous slopes, crucial for understanding and working with derivatives effectively.

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Most popular questions from this chapter

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

A challenging derivative Find \(\frac{d y}{d x},\) where \(\sqrt{3 x^{7}+y^{2}}=\sin ^{2} y+100 x y\).

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of $$f(x)=x^{3} ;(8,2)$$

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}$$

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x}\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this implies that the rule is true for \(k=n+1 .\) (Hint: Write \(e^{(n+1) x}\) as the product of two functions, and use the Product Rule.)

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