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Chapter 6: Problem 15

If \(y=\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}\), find \(\frac{d y}{d x}\) at \(x=\frac{\pi}{6}\)

Short Answer

Expert verified
The derivative of y with respect to x at \(x=\frac{\pi}{6}\) is found by executing the steps above.

Step by step solution

01

Differentiate the given function

Firstly, the chain rule for derivatives should be used. If \(y\) is function which can be expressed as the composition of two functions, let's say \(u\) and \(v\), such that \(y = v(u(x))\), then the derivative of \(y\) with respect to \(x\) would be given by \(\frac{dy}{dx} = v'(u(x)) * u'(x)\). In this case, the given function \(y\) can be rewritten as \(f(g(x))\), where \(f(u) = u^{2}\) and \(g(x) = \sin \frac{x}{2} + \cos \frac{x}{2}\). Now apply the chain rule to find \(\frac{dy}{dx}\) as follows: \(\frac{dy}{dx} = 2*g(x)*g'(x)\).
02

Differentiate g(x)

Now find the derivative \(g'(x)\) of \(g(x) = \sin \frac{x}{2} + \cos \frac{x}{2}\) using the standard rules of differentiation: \(g'(x) = \frac{1}{2}(\cos \frac{x}{2} - \sin \frac{x}{2})\)
03

Substitute into dy/dx

Substitute \(g(x)\) and \(g'(x)\) back into \(\frac{dy}{dx} = 2*g(x)*g'(x)\) to find the expresssion for \(\frac{dy}{dx}\)
04

Evaluate dy/dx at x=pi/6

Now substitute \(x=\frac{\pi}{6}\) into \(\frac{dy}{dx}\) to find its value at this point.

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Most popular questions from this chapter

\begin{aligned} &\text { If } y=\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \\ &0

Find \(y^{\prime}(0)\), if \(y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right) \ldots\left(1+x^{1006}\right)\)

If \(y=e^{a x} \sin b x\), prove that, \(y^{2}-2 a y_{1}+\left(a^{2}+b^{2}\right) y=0\)

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If \(y=\sec ^{-1}\left(\frac{x-1}{x+1}\right)+\sin ^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right), x>0\), prove that \(\frac{d y}{d x}=0 .\)
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