Question

d/dx(x/2+1/2log|sinx+cosx|)
differentiate it​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

◾The given function is

$y = \frac{x}{2}+\frac{1}{2} \times \ log |sinx + cosx |$

◾Differentiate the function wrt x

$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}+\frac{1}{2} \times \ log |sinx + cosx |)$

$=(\frac{1}{2}+\frac{1}{2} \times( \frac{1}{sinx + cosx}) (cosx - sinx))$

$= (\frac{1}{2}+\frac{cosx - sinx}{2(sinx + cosx)})$

$= \frac{2(sinx + cosx) + 2(cosx - sinx)}{4(sinx + cosx) }$

$=\frac{2sinx + 2 cosx +2 cosx - 2sinx}{4 sinx + 4 cosx }$

$= \frac{ 4cosx}{4sinx + 4cosx}$

$\frac{dy}{dx}=\frac{cosx}{sinx + cosx}$

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⚫Derivatives of some standard composite functions :

1] y =[ f(x) ]^n

➡ dy / dx = n[f(x)]^(n - 1)x f'(x )

2] y = sin f (x)

➡dy / dx = cos f (x) x f' (x)

3] y = √ ( f (x ) )

➡dy / dx =( 1 / 2 √[ f(x)]) x f'(x )

4] y = k

➡ dy / dx = 0 [ Note : it is not a composite function ]

5] y = cos f (x )

➡ dy /dx = -sin f (x) x f'(x )

6] y = e^([f (x) ])

➡dy /dx = e^([f (x) ]) x f'(x)

7] y = a^[f(x)]

➡ dy /dx = a^[ f (x) ] x log a x f'(x)

8] y = log [f (x) ]

➡dy /dx =( 1 / f (x) ) x [f'(x) ]

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