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Find \(\frac{d y}{d x}\) if \(y=\cos x^{2}\).

Short Answer

Expert verified
The derivative of the function \(y=\cos x^{2}\) with respect to 'x' is \(\frac{dy}{dx} = -2x\sin x^2\).

Step by step solution

01

Identify the Inner Function

Here, \(x^2\) is the inner function. Let's denote it as 'u', therefore \(u = x^2\).
02

Differentiate the Inner Function

Differentiate 'u' with respect to 'x'. The derivative of \(u=x^2\) is \(u' = 2x\).
03

Identify and Differentiate the Outer Function

The outer function is the cos function, which is applied to 'u'. Therefore, we have \(y = \cos u\). The derivative of cosine is negative sine. So, \(y'\) or \(\frac{dy}{du}\) = \(-\sin u\). Note that we are differentiating with respect to 'u', not 'x'.
04

Apply the Chain Rule

According to the chain rule, when we have a composite function such as this, the derivative is given as the product of the derivative of the outer function and the derivative of the inner function. In mathematical terms, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). Substitute the values of \(\frac{dy}{du}\) and \(\frac{du}{dx}\) and we have \(\frac{dy}{dx} = -\sin u \cdot 2x\)
05

Substitute Back the Original Function

Now, we know that \(u = x^2\). Substitute 'u' with \(x^2\) in the final expression for the derivative. So, \(\frac{dy}{dx} = -2x\sin x^2\).

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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 fundamental concept in calculus used to differentiate composite functions. To apply the chain rule, you need to identify both the inner and outer functions in an expression.

Consider a function \[ y = f(g(x)) \] where \( g(x) \) is the inner function, and \( f(u) \) is the outer function with \( u = g(x) \). This is a chain of functions. The chain rule states that the derivative of this composite function is:
  • \( \frac{dy}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \)
In simpler terms, you first differentiate the outer function with respect to the inner function and then multiply it by the derivative of the inner function with respect to the original variable \( x \). This rule helps in breaking down complex derivatives into manageable steps. Remember, the power of the chain rule lies in its ability to handle layers of functions nested within each other effectively.

When practicing, always clearly separate and identify the inner and outer functions before applying the rule.
Composite Functions
Composite functions consist of applying one function to the results of another function. They are written as \(f(g(x))\), where \(x\) is first input into \(g\), and then \(g(x)\) becomes the input for \(f\).

In mathematical terms, if you have:
  • \(y = f(g(x))\)
To break it down:
  • First, find \(g(x)\) which is referred to as the inner function.
  • Then, apply \(f\), the outer function, to \(g(x)\).
Understanding composite functions is key because they often appear in calculus problems involving the chain rule. For example, in the problem \(y = \cos(x^2)\), \(x^2\) is an expression inside the \(\cos\) function. Thus, \(x^2\) is the inner function, and \(\cos(u)\) is the outer function.

When dealing with composite functions, a clear identification of the components helps navigate through the solution effectively. By analyzing each function layer individually, you can apply appropriate differentiation strategies like the chain rule easily.
Trigonometric Derivatives
Trigonometric derivatives are essential for differentiating functions involving trigonometric ratios like sine, cosine, and tangent.

Each trigonometric function has a specific derivative:
  • 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 given exercise, the function \(y = \cos(x^2)\) requires using the derivative \(-\sin(x)\). However, remember that you are actually finding the derivative with respect to \(u\) first when applying the chain rule. That's why the derivative part becomes \(-\sin(u)\), where \(u = x^2\).

After this step, we take into account the inner differentiation, resulting in the final derivative. Trigonometric derivatives often intertwine with the application of the chain rule when composite functions are involved. Therefore, knowing these derivatives by heart increases your speed and accuracy when working through complex calculations.

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Most popular questions from this chapter

\(\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{6}+h\right)-\tan \left(\frac{\pi}{6}\right)}{h}=\) (A) \(\frac{4}{3}\) (B) \(\sqrt{3}\) (C) 0 (D) \(\frac{3}{4}\)

Water is draining at the rate of 48\(\pi \mathrm{f}^{3} / \mathrm{second}\) from the vertex at the bottom of a conical tank whose diameter at its base is 40 feet and whose height is 60 feet. (a) Find an expression for the volume of water in the tank, in terms of its radius, at the surface of the water. (b) At what rate is the radius of the water in the tank shrinking when the radius is 16 feet? (c) How fast is the height of the water in the tank dropping at the instant that the radius is 16 feet?

If \(f(x)=\left\\{\begin{array}{l}{x^{2}+5 \text { if } x<2} \\ {7 x-5 \text { if } x \geq 2}\end{array}, \text { for all real numbers } x, \text { which of the following must be true? }\right.\) I. \(f(x)\) is continuous everywhere. II. \(f(x)\) is differentiable everywhere. III. \(f(x)\) has a local minimum at \(x=2\) (A) I only (B) I and II only (C) II and III only (D) I, II, and III

Consider the equation \(x^{2}-2 x y+4 y^{2}=64\) (a) Write an expression for the slope of the curve at any point \((x, y)\). (b) Find the equation of the tangent lines to the curve at the point \(x=2\). (c) Find \(\frac{d^{2} y}{d x^{2}}\) at \((0,4)\).

Find the area of the region between the two curves in each problem, and be sure to sketch each one. (We gave you only endpoints in one of them) The answers are in Chapter 19 . The curve \(y=x^{2}\) and the curve \(y=4 x-x^{2}\).

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