Question

Find the derivative of the following functions from first principle:
$cos(x - \frac{\pi}{8} )$
​

Answer 1

$\boxed{\sf \red{ Answer}}$

$\mathsf{\sqrt{2} cosx}</p><p>$

$\boxed{\sf \red{ Explanation}}$

To Find:

Value of,

$\mathsf{\pink{cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }}$

Solution:

$\mathsf{ cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }</p><p>$

W.k.t

$\boxed{\bold\blue{ cos(A+B) = cosA.cosB - sinA.SinB}}$

$\boxed{\bold\blue{ cos(A-B) = cosA.cosB + sinA.SinB}}$

By using the identities

$\mathsf{\implies cos(\dfrac{\pi}{4}+x) + cos (\dfrac{\pi}{4}-x) }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx - sin \dfrac{\pi}{4}. sinx + cos\dfrac{\pi}{4} cosx +sin \dfrac{\pi}{4}. sinx }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx - \cancel{sin \dfrac{\pi}{4}. sinx }+ cos\dfrac{\pi}{4} cosx +\cancel{sin \dfrac{\pi}{4}. sinx} }</p><p>$

$\mathsf{\implies cos\dfrac{\pi}{4} cosx + cos\dfrac{\pi}{4} cosx }</p><p>$

$\mathsf{\implies 2(cos\dfrac{\pi}{4} cosx ) }</p><p>$

$\mathsf{\implies 2( \dfrac{1}{\sqrt{2}} cosx ) }</p><p>$

$\mathsf{\implies{\sqrt{2} \times \sqrt{2} ( \dfrac{1}{\sqrt{2}} cosx ) }}</p><p>$

$\mathsf{\implies{\sqrt{2} \times \cancel{\sqrt{2}} ( \dfrac{1}{\cancel{\sqrt{2}}} cosx ) }}</p><p>$

$\mathsf{\implies \sqrt{2} cosx }</p><p>$

#$\texttt{\red{ Marzi} \: \blue{Here}...!}$

Answer 2

Answer:

Let f(x)=cos(x−

8

π

)

Thus using first principle

f

′

(x)=

x→0

lim

h

f(x+h)−f(x)

=

x→0

lim

h

1

[cos(x+h−

8

π

)−cos(x−

8

π

)]

=

x→0

lim

h

1

[−2sin

2

(x+h−

8

π

+x−

8

π

)

sin(

2

x+h−

8

π

−x+

8

π

)]

=

x→0

lim

h

1

[−2sin(

2

2x+h−

4

π

)sin

2

h

]

=

x→0

lim

[−sin(

2

2x+h−

4

π

)

(

2

h

)

sin(

2

h

)

]

=−sin(

2

2x+0−

4

π

)⋅1=−sin(x−

8

π

)

Step-by-step explanation:

PN MAJHI JAGA DUSRYA KUNALA DILAS TR BG TU