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Chapter 3: Problem 8

Evaluate the derivatives of the following functions. $$f(x)=x \sin ^{-1} x$$

Short Answer

Expert verified
Answer: The derivative of the function $$f(x) = x \sin^{-1}(x)$$ with respect to x is $$f'(x) = \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$$.

Step by step solution

01

Identify the Chain Rule Components

In the given function $$f(x) = x \sin^{-1}(x)$$, there are two components: $$u(x) = x$$ and $$v(x) = \sin^{-1}(x)$$. We need to find the derivatives of both these components and then apply the chain rule to find the derivative of f(x).
02

Find the derivative of x

The derivative of $$u(x) = x$$ with respect to x is $$u'(x) = 1$$.
03

Find the derivative of sin^{-1}(x)

To find the derivative of $$v(x) = \sin^{-1}(x)$$, apply the formula for the derivative of an inverse sine function: $$\frac{d}{dx} \sin^{-1}(x)= \frac{1}{\sqrt{1-x^2}}$$. So, the derivative of $$v(x)$$ is $$v'(x) = \frac{1}{\sqrt{1-x^2}}$$.
04

Apply the Chain Rule

Now, apply the chain rule to find the derivative of $$f(x) = x \sin^{-1}(x)$$. Since $$f(x) = u(x)v(x)$$, the chain rule states that its derivative concerning x is given by: $$f'(x) = u'(x)v(x) + u(x)v'(x)$$
05

Compute the derivative of f(x) concerning x

Using the previously calculated derivatives, we can write the derivative of f(x) as follows: $$f'(x) = u'(x)v(x) + u(x)v'(x) = 1 \cdot \sin^{-1}(x) + x \cdot \frac{1}{\sqrt{1-x^2}}$$
06

Simplify the expression

The final step is to simplify the expression for the derivative of f(x) concerning x: $$f'(x) = \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$$ The derivative of the given function $$f(x) = x \sin^{-1}(x)$$ is $$f'(x) = \sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}$$.

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Most popular questions from this chapter

A challenging derivative Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\).

Derivatives from tangent lines Suppose the line tangent to the graph of \(f\) at \(x=2\) is \(y=4 x+1\) and suppose \(y=3 x-2\) is the line tangent to the graph of \(g\) at \(x=2 .\) Find an equation of the line tangent to the following curves at \(x=2\) a. \(y=f(x) g(x)\) b. \(y=\frac{f(x)}{g(x)}\)

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=3 x-4$$

Visualizing tangent and normal lines a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises \(63-68 .)\) b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right);\) \(\left(x_{0}, y_{0}\right)=(2,2)\) (lima莽on of Pascal)

Means and tangents Suppose \(f\) is differentiable on an interval containing \(a\) and \(b,\) and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x},\) show that \(c=\sqrt{a b},\) the geometric mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a>0\) and \(b>0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.
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