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Chapter 6: Problem 21

If \(y=\log (\sin x+\cos x)\), find \(\frac{d y}{d x}\)

Short Answer

Expert verified
\(\frac{d y}{d x} = \frac{\cos x - \sin x}{\sin x + \cos x}\)

Step by step solution

01

Identify the type of function

The function is in the form of \(y = \log(f(x))\), where \(f(x) = \sin x + \cos x\).
02

Apply chain rule

According to the chain rule, the derivative of a composite function is the derivative of the outside function times the derivative of the inner function. Hence \(\frac{d y}{d x} = \frac{1}{f(x)} \cdot f'(x)\).
03

Derivative of inner function

The derivative of the inner function \(f(x) = \sin x + \cos x\) is \(f'(x) = \cos x - \sin x\). This comes from the derivative of \(\sin x\) being \(\cos x\) and the derivative of \(\cos x\) being \(-\sin x\).
04

Substitute f'(x) and f(x) in the equation

Substitute \(f'(x)\) and \(f(x)\) into the equation from step 2, we get \(\frac{d y}{d x} = \frac{1}{\sin x + \cos x} \cdot (\cos x - \sin x)\).

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Most popular questions from this chapter

For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)For the function \(f(x)=\left[x+\frac{1}{2}\right]+\left[x-\frac{1}{2}\right]-[2 x]\), Rolle's theorem is applicable on the interval (a) \(\left[0, \frac{1}{4}\right]\) (b) \(\left[\frac{1}{4}, 1\right]\) (c) \(\left[0, \frac{1}{2}\right]\) (d) \([0,1]\)

If \(x^{m} y^{n}=(x+y)^{m+n}\), prove that, \(\frac{d y}{d x}=\frac{y}{x}\)

Suppose \(f\) is a differentiable function such that \(f(g(x))=x\) and \(f^{\prime}(x)=1+(f(x))^{2}\), then prove that \(g^{\prime}(x)=\frac{1}{1+x^{2}}\)

Find \(y^{\prime}(0)\), if\(y=(x+1)(x+2)(x+3) \ldots(x+2012)\)

Rolle's theorem is applicable for the function \(f(x)=(x-1)|x|+|x-1|\) in the interval (a) \([0,1]\) (b) \(\left[\frac{1}{4}, \frac{3}{4}\right]\) (c) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (d) \(\left[\frac{1}{5}, \frac{6}{7}\right]\)
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