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Tangent lines Find an equation of the line tangent to \(y=x^{\sin x}\) at the point \(x=1\)

Short Answer

Expert verified
Answer: The equation of the tangent line to the curve \(y = x^{\sin x}\) at the point \(x = 1\) is \(y = (\cos 1 + \sin 1)(x - 1) + 1\).

Step by step solution

01

Find the derivative of the function \(y = x^{\sin x}\)

To find the derivative of the function \(y = x^{\sin x}\), we will use the chain rule and the exponential rule. Let \(g(x) = x^{\sin x}\) and \(h(x) = \sin x\). Then, we can rewrite \(y\) as: \(y(x) = g^{(h(x))}\) By applying the chain rule, we get: \(y'(x) = g'(h(x)) \cdot h'(x)\) Taking the derivative of \(h(x) = \sin x\) with respect to \(x\), \(h'(x) = \cos x\) Now, we will find the derivative of \(g(x) = x^{\sin x}\) with respect to \(x\) using the natural logarithm to simplify the process: \(\ln(g(x)) = \sin x \ln x\) Taking the derivative with respect to \(x\), \(\frac{g'(x)}{g(x)} = \frac{\cos x}{x \sin x} + \sin x \frac{1}{x} \frac{1}{\sin x}\) \(g'(x) = (x^{\sin x})\left(\frac{\cos x}{x \sin x} + \frac{\sin x}{x}\right)\) Thus, our derivative is: \(y'(x) = g'(x) = (x^{\sin x})\left(\frac{\cos x}{x \sin x} + \frac{\sin x}{x}\right)\)
02

Find the slope of the tangent line at the given point: \(x = 1\)

Now, we will find the slope of the tangent line at the given point \(x = 1\) by substituting the given value of \(x\) into the derivative: \(y'(1) = (1^{\sin 1})\left(\frac{\cos 1}{1 \sin 1} + \frac{\sin 1}{1}\right) = \cos 1 + \sin 1\) Therefore, the slope of the tangent line at the given point is \(\cos 1 + \sin 1\).
03

Find the coordinates of the point on the curve where \(x = 1\)

To find the coordinates of the point on the curve where \(x = 1\), we substitute this value into the original function \(y = x^{\sin x}\): \(y(1) = 1^{\sin 1} = 1\) So, the coordinates of the point are \((1, 1)\).
04

Find the equation of the tangent line using point-slope form

Finally, we will use the point-slope form of a line to find the equation of the tangent line. The point-slope form is: \(y - y_1 = m(x - x_1)\) Substituting our values for the point \((x_1, y_1) = (1, 1)\) and slope \(m = \cos 1 + \sin 1\), we get: \(y - 1 = (\cos 1 + \sin 1)(x - 1)\) Therefore, the equation of the tangent line to the curve \(y = x^{\sin x}\) at the point \(x = 1\) is: \(y = (\cos 1 + \sin 1)(x - 1) + 1\)

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Key Concepts

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

Derivative of Exponential Functions
Understanding the derivative of exponential functions is a fundamental concept in calculus. An exponential function can be generally represented as \(a^{x}\), where \(a\) is a constant and \(x\) is the exponent that varies. To differentiate such functions, we often need to use logarithmic differentiation when the exponent itself is a function of \(x\), which is the case in our exercise with \(y = x^{\sin x}\).

The technique involves taking the natural logarithm, denoted as \(\ln\), of both sides of the equation to bring the exponent down. Then the derivative is obtained with respect to \(x\), applying standard rules of differentiation. It's important to remember that the derivative of the exponential function \(e^{x}\) is unique because it is its own derivative. For other bases, a conversion is often applied to relate it to the natural exponential \(e\).
Chain Rule in Differentiation
The chain rule is a powerful tool in calculus used to differentiate composite functions. A composite function is a function composed of two or more functions, such that one function is applied to the result of another function. The formal definition involves \(f(g(x))\), where \(f\) and \(g\) are both functions of \(x\).

When differentiating a composite function, the chain rule states that the derivative of \(f(g(x))\) is \(f'(g(x))\) times \(g'(x)\). This rule allows us to break down more complex expressions into simpler parts that can be differentiated individually, then multiplied together. The chain rule was applied in our exercise by first differentiating the outer function and then multiplying it by the derivative of the inner function.
Point-Slope Form
The point-slope form is a convenient way to write the equation of a line when you know a point on the line and its slope. The general formula is \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope of the line, and \(x_1, y_1)\) represents the coordinates of the known point. This form is particularly useful for quickly writing the equation of a tangent line to a curve at a specific point, as demonstrated in our exercise.

To establish the point-slope form, you need the derivative of the function at the point of tangency to determine the slope, and then you apply the coordinates of the point directly into the formula. Adjusting the equation to solve for \(y\) gives you a linear equation that represents the tangent line.
Implicit Differentiation
Implicit differentiation is a technique used when it is difficult or impossible to solve an equation for one variable in terms of the other variable before differentiating. This usually occurs with functions that are defined implicitly rather than explicitly. With implicit differentiation, you differentiate both sides of the equation with respect to \(x\), treating \(y\) as an implicit function of \(x\), and use the chain rule as necessary. After differentiating, we often solve for \(y'\), which represents the derivative of \(y\) with respect to \(x\).

In scenarios where we have \(y\) as a function raised to a power of another function of \(x\), as in \(y = x^{\sin x}\), implicit differentiation enables us to find the derivative \(y'(x)\) without the need to isolate \(y\) first. This is applicable in more complicated situations where expressing \(y\) explicitly in terms of \(x\) is not straightforward or possible.

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Most popular questions from this chapter

A mixing tank A 500 -liter (L) tank is filled with pure water. At time \(t=0,\) a salt solution begins flowing into the tank at a rate of \(5 \mathrm{L} / \mathrm{min.}\) At the same time, the (fully mixed) solution flows out of the tank at a rate of \(5.5 \mathrm{L} / \mathrm{min}\). The mass of salt in grams in the tank at any time \(t \geq 0\) is given by $$M(t)=250(1000-t)\left(1-10^{-30}(1000-t)^{10}\right)$$ and the volume of solution in the tank is given by $$V(t)=500-0.5 t$$ a. Graph the mass function and verify that \(M(0)=0\) b. Graph the volume function and verify that the tank is empty when \(t=1000 \mathrm{min}\) c. The concentration of the salt solution in the tank (in \(\mathrm{g} / \mathrm{L}\) ) is given by \(C(t)=M(t) / V(t) .\) Graph the concentration function and comment on its properties. Specifically, what are \(C(0)\) \(\underset{t \rightarrow 1000^{-}}{\operatorname{and}} C(t) ?\) d. Find the rate of change of the mass \(M^{\prime}(t),\) for \(0 \leq t \leq 1000\) e. Find the rate of change of the concentration \(C^{\prime}(t),\) for \(0 \leq t \leq 1000\) f. For what times is the concentration of the solution increasing? Decreasing?

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$\left(x^{2}+y^{2}\right)^{2}=\frac{25}{4} x y^{2} ;(1,2)$$ (Graph cant copy)

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no point of intersection (for \(x>0\) ), while the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 < p < 3,\) the graphs of \(y=x^{p}\) and \(y=e^{x}\) have exactly one point of intersection (for \(x > 0) .\) Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

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