The question
Show that the function f ( x ) = 16 sin x 4 + cos x − x f(x) = \dfrac{16\sin x}{4 + \cos x} - x f ( x ) = 4 + cos x 16 sin x − x ( π 2 , π ) \left(\dfrac{\pi}{2}, \pi\right) ( 2 π , π )
Exam answer Show all explanations
Using the quotient rule,
f ′ ( x ) = 16 ( 4 + cos x ) cos x ( 4 + cos x ) 2 + 16 sin 2 x ( 4 + cos x ) 2 − 1 \begin{aligned} f'(x) &= \frac{16(4 + \cos x)\cos x}{(4 + \cos x)^2} \\ &\quad + \frac{16\sin^2x}{(4 + \cos x)^2} - 1 \end{aligned} f ′ ( x ) = ( 4 + cos x ) 2 16 ( 4 + cos x ) cos x + ( 4 + cos x ) 2 16 sin 2 x − 1 Why?
The first term is a quotient with u = 16 sin x u = 16\sin x u = 16 sin x v = 4 + cos x v = 4 + \cos x v = 4 + cos x
d d x ( u v ) = v u ′ − u v ′ v 2 \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{vu' - uv'}{v^2} d x d ( v u ) = v 2 v u ′ − u v ′ Here u ′ = 16 cos x u' = 16\cos x u ′ = 16 cos x v ′ = − sin x v' = -\sin x v ′ = − sin x − u v ′ = + 16 sin 2 x -uv' = +16\sin^2x − u v ′ = + 16 sin 2 x
v u ′ = ( 4 + cos x ) ( 16 cos x ) = 16 ( 4 + cos x ) cos x u v ′ = ( 16 sin x ) ( − sin x ) = − 16 sin 2 x − u v ′ = + 16 sin 2 x \begin{aligned} vu' &= (4 + \cos x)(16\cos x) \\ &= 16(4 + \cos x)\cos x \\[4pt] uv' &= (16\sin x)(-\sin x) \\ &= -16\sin^2x \\[4pt] -uv' &= +16\sin^2x \end{aligned} v u ′ u v ′ − u v ′ = ( 4 + cos x ) ( 16 cos x ) = 16 ( 4 + cos x ) cos x = ( 16 sin x ) ( − sin x ) = − 16 sin 2 x = + 16 sin 2 x Split v u ′ − u v ′ v 2 \frac{vu' - uv'}{v^2} v 2 v u ′ − u v ′ v 2 = ( 4 + cos x ) 2 v^2 = (4 + \cos x)^2 v 2 = ( 4 + cos x ) 2
The derivative of x x x − 1 -1 − 1
Using cos 2 x + sin 2 x = 1 \cos^2x + \sin^2x = 1 cos 2 x + sin 2 x = 1
f ′ ( x ) = 64 cos x + 16 ( 4 + cos x ) 2 − 1 \begin{aligned} f'(x) &= \frac{64\cos x + 16}{(4 + \cos x)^2} - 1 \end{aligned} f ′ ( x ) = ( 4 + cos x ) 2 64 cos x + 16 − 1 Why?
The two fractions have the same denominator, so add their numerators and expand:
16 ( 4 + cos x ) cos x + 16 sin 2 x = 64 cos x + 16 cos 2 x + 16 sin 2 x = 64 cos x + 16 \begin{aligned} &16(4 + \cos x)\cos x \\ &\quad + 16\sin^2x \\ &= 64\cos x + 16\cos^2x \\ &\quad + 16\sin^2x \\ &= 64\cos x + 16 \end{aligned} 16 ( 4 + cos x ) cos x + 16 sin 2 x = 64 cos x + 16 cos 2 x + 16 sin 2 x = 64 cos x + 16 The last step uses 16 cos 2 x + 16 sin 2 x = 16 16\cos^2x + 16\sin^2x = 16 16 cos 2 x + 16 sin 2 x = 16
16 ( 4 + cos x ) cos x = 16 ( 4 ) cos x + 16 cos x cos x = 64 cos x + 16 cos 2 x 16 cos 2 x + 16 sin 2 x = 16 ( cos 2 x + sin 2 x ) = 16 ( 1 ) = 16 \begin{aligned} &16(4 + \cos x)\cos x \\ &= 16(4)\cos x + 16\cos x\cos x \\ &= 64\cos x + 16\cos^2x \\[4pt] &16\cos^2x + 16\sin^2x \\ &= 16(\cos^2x + \sin^2x) \\ &= 16(1) = 16 \end{aligned} 16 ( 4 + cos x ) cos x = 16 ( 4 ) cos x + 16 cos x cos x = 64 cos x + 16 cos 2 x 16 cos 2 x + 16 sin 2 x = 16 ( cos 2 x + sin 2 x ) = 16 ( 1 ) = 16 f ′ ( x ) = 56 cos x − cos 2 x ( 4 + cos x ) 2 = cos x ( 56 − cos x ) ( 4 + cos x ) 2 \begin{aligned} f'(x) &= \frac{56\cos x - \cos^2x}{(4 + \cos x)^2} \\ &= \frac{\cos x\,(56 - \cos x)}{(4 + \cos x)^2} \end{aligned} f ′ ( x ) = ( 4 + cos x ) 2 56 cos x − cos 2 x = ( 4 + cos x ) 2 cos x ( 56 − cos x ) Why?
Write 1 over the same denominator and subtract:
1 = ( 4 + cos x ) 2 ( 4 + cos x ) 2 1 = \frac{(4 + \cos x)^2}{(4 + \cos x)^2} 1 = ( 4 + cos x ) 2 ( 4 + cos x ) 2 Expand with ( a + b ) 2 = a 2 + 2 a b + b 2 (a + b)^2 = a^2 + 2ab + b^2 ( a + b ) 2 = a 2 + 2 ab + b 2
( 4 + cos x ) 2 = 4 2 + 2 ( 4 ) ( cos x ) + cos 2 x = 16 + 8 cos x + cos 2 x \begin{aligned} &(4 + \cos x)^2 \\ &= 4^2 + 2(4)(\cos x) + \cos^2x \\ &= 16 + 8\cos x + \cos^2x \end{aligned} ( 4 + cos x ) 2 = 4 2 + 2 ( 4 ) ( cos x ) + cos 2 x = 16 + 8 cos x + cos 2 x 64 cos x + 16 − ( 16 + 8 cos x + cos 2 x ) = 64 cos x + 16 − 16 − 8 cos x − cos 2 x = 56 cos x − cos 2 x \begin{aligned} &64\cos x + 16 \\ &\quad - (16 + 8\cos x + \cos^2x) \\ &= 64\cos x + 16 - 16 \\ &\quad - 8\cos x - \cos^2x \\ &= 56\cos x - \cos^2x \end{aligned} 64 cos x + 16 − ( 16 + 8 cos x + cos 2 x ) = 64 cos x + 16 − 16 − 8 cos x − cos 2 x = 56 cos x − cos 2 x (64 cos x − 8 cos x = 56 cos x 64\cos x - 8\cos x = 56\cos x 64 cos x − 8 cos x = 56 cos x 16 − 16 = 0 16 - 16 = 0 16 − 16 = 0
Then take out the common factor cos x \cos x cos x
56 cos x − cos 2 x = cos x ( 56 ) − cos x ( cos x ) = cos x ( 56 − cos x ) \begin{aligned} &56\cos x - \cos^2x \\ &= \cos x(56) - \cos x(\cos x) \\ &= \cos x\,(56 - \cos x) \end{aligned} 56 cos x − cos 2 x = cos x ( 56 ) − cos x ( cos x ) = cos x ( 56 − cos x ) In ( π 2 , π ) \left(\frac{\pi}{2}, \pi\right) ( 2 π , π ) cos x < 0 \cos x < 0 cos x < 0 56 − cos x > 0 56 - \cos x > 0 56 − cos x > 0 ( 4 + cos x ) 2 > 0 (4 + \cos x)^2 > 0 ( 4 + cos x ) 2 > 0
Why?
For x x x π 2 \frac{\pi}{2} 2 π π \pi π − 1 < cos x < 0 -1 < \cos x < 0 − 1 < cos x < 0
So 56 − cos x 56 - \cos x 56 − cos x 4 + cos x 4 + \cos x 4 + cos x
So f ′ ( x ) < 0 f'(x) < 0 f ′ ( x ) < 0 ( π 2 , π ) \left(\frac{\pi}{2}, \pi\right) ( 2 π , π )
Why?
The sign of f ′ ( x ) f'(x) f ′ ( x )
A function whose derivative is negative at every point of an interval is strictly decreasing on that interval.
∴ f ( x ) \therefore\ f(x) ∴ f ( x ) ( π 2 , π ) \left(\dfrac{\pi}{2}, \pi\right) ( 2 π , π )
Where this comes from Chapter 6, Application of Derivatives : Strictly decreasing function on an interval.
Question type: Show that a function is increasing everywhere
How to spot it: click to see Type 4. Show that a function is increasing everywhere
Spot it: "Show that (or prove that) f f f R \R R
Here: Here the set is ( π 2 , π ) \left(\frac{\pi}{2}, \pi\right) ( 2 π , π ) R \mathbb{R} R f ′ ( x ) = cos x ( 56 − cos x ) ( 4 + cos x ) 2 f'(x) = \frac{\cos x\,(56 - \cos x)}{(4 + \cos x)^2} f ′ ( x ) = ( 4 + c o s x ) 2 c o s x ( 56 − c o s x ) cos x \cos x cos x f ′ ( x ) f'(x) f ′ ( x )
Quick lesson · Worked NCERT solutions · Interactive lesson