Question

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

Short answer

Answer: The equation of the tangent line is \(y = (\sin 1)(x-1) + 1^{\sin 1}\).

Step-by-step solution

Find the derivative of y with respect to x

To find the derivative of \(y = x^{\sin x}\) with respect to x, we will need to use the chain rule and power rule. First, we notice that \(y = u^v\) where \(u = x\) and \(v = \sin x\). Let's find the derivatives of \(u\) and \(v\) with respect to x: \(\frac{du}{dx} = \frac{d(x)}{dx} = 1\) \(\frac{dv}{dx} = \frac{d(\sin x)}{dx} = \cos x\) Now, we need to apply the chain rule: \(\frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du} + \frac{dv}{dx} \cdot \frac{dy}{dv}\) We will find \(\frac{dy}{du}\) and \(\frac{dy}{dv}\) using the power rule: \(\frac{dy}{du} = \frac{d(u^v)}{du} = vu^{v-1}\) \(\frac{dy}{dv} = \frac{d(u^v)}{dv} = u^v \ln u\) Now we plug our values back into the chain rule: \(\frac{dy}{dx} = 1 \cdot (\sin x)(x^{\sin x - 1})+(\cos x)(x^{\sin x}\ln x)\)

Evaluate the derivative at x=1

To find the slope of the tangent line, we need to plug x=1 into our derived formula: \(\frac{dy}{dx}|_{x=1} = (\sin 1)(1^{\sin1 - 1})+(\cos 1)(1^{\sin1}\ln 1)\) Since \(\ln 1 = 0\), the second term becomes 0, so : \(\frac{dy}{dx}|_{x=1} = \sin 1\)

Find the equation of the tangent line

Now that we have the slope, we can find the equation of the tangent line using the point-slope form of a linear equation: \(y - y_1 = m(x - x_1)\) Where m = slope of the tangent line = \(\sin 1\), \(x_1 = 1\), and \(y_1 = 1^{\sin 1}\). Plugging in the values, \(y - 1^{\sin 1} = (\sin 1)(x - 1)\) So the equation of the line tangent to \(y = x^{\sin x}\) at the point \(x=1\) is: \(y = (\sin 1)(x-1) + 1^{\sin 1}\)

Key concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is essential in calculus, especially when dealing with complex functions involving exponents that are not just constants. In general, if you have an exponential function of the form \(y = a^x\), where \(a\) is a constant, the derivative with respect to \(x\) is \(y' = a^x \ln(a)\). However, the function \(y = x^{\sin x}\) introduces a variable in the exponent, which cannot be differentiated using the standard formula for exponential functions. Instead, we must employ a combination of calculus rules to find its derivative.

Chain Rule Calculus

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. If you have a function \(y = f(g(x))\), where \(f\) and \(g\) are both differentiable, the chain rule states that the derivative \(y' = f'(g(x))g'(x)\). When applied to the function \(y = x^{\sin x}\), we treat \(x\) and \(\sin x\) as separate functions that compose our main function. The chain rule allows us to differentiate \(y\) with respect to \(x\) by also considering the derivatives of \(x\) and \(\sin x\), thus breaking down a complex derivative into more manageable parts.

Power Rule Calculus

The power rule is another important concept in calculus, which states that for any function \(y = x^n\), where \(n\) is a real number, the derivative is \(y' = nx^{n-1}\). This rule makes it straightforward to find derivatives of functions with power expressions. However, in the case of \(y = x^{\sin x}\), we see that the exponent itself is a function of \(x\). Hence, we combine the power rule with the chain rule to find the derivative, taking into account that the exponent is not constant but rather a variable function dependent on \(x\).

Slope of Tangent Line

The slope of the tangent line to a curve at a particular point is the value of the derivative of the function at that point. This slope indicates the steepness of the line and is essential in constructing the line tangent to the curve. For instance, in the given exercise, we find the slope of the tangent line to \(y = x^{\sin x}\) at \(x = 1\) by evaluating the derivative at that point, resulting in \(m = \sin(1)\). The point-slope form, \(y - y_1 = m(x - x_1)\), is then used to write the equation of the tangent line, using the slope and the coordinates of the given point on the curve.