Question

Find the derivative of cos ax by using the first principle method

Answer 1

Step-by-step explanation:

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Answer 2

cos ax derivative by first principle method is - a sin ax.

Step-by-step explanation:

Given:

cos ax

To Find:

cos ax derivative by first principle method.

Formula Used:

Function $y= f(x)$

Derivative of f(x) by first principle

$f'(x)= \lim_{h \to \ 0 } \frac{f(x+h)-f(h)}{h}$    ------------- formula no. 01

Solution:

As given- cos ax

$f(x) =cosax \\f(x+h) = cos a( x+h) = cos( ax+ah)$  

Applying the formula no. 01.

$f'(x)= \lim_{h \to \ 0 } \frac{cos(ax+ah)- cos ax}{h}$

        $= \lim_{h \to \ 0 } \frac{-2sin(\frac{ax+ah+ax}{2} ) sin(\frac{ax+ah-ax}{2} )}{h}$

       $= \lim_{h \to \ 0 } \frac{-2sin(\frac{2ax+ah}{2} ) sin(\frac{ah}{2} )}{h}$

      $= \lim_{h \to \ 0 } -2 sin( ax+\frac{ah}{2} ) . \lim_{h \to \ 0} \frac{sin ( \frac{ah}{2} )}{h}$

      $= -2 \lim_{h \to \ 0 } sin( ax+\frac{ah}{2} ) . \lim_{h \to \ 0} \frac{sin ( \frac{ah}{2} )}{h}$

     $= \lim_{h \to \ 0 } -2 sin( ax+\frac{ah}{2} ) . \lim_{h \to \ 0} \frac{sin ( \frac{ah}{2} )}{\frac{ah}{2} \times \frac{2}{a} }$

    $= \lim_{h \to \ 0 } -2 sin( ax+\frac{ah}{2} ) . \frac{a}{2} \lim_{h \to \ 0} \frac{sin ( \frac{ah}{2} )}{\frac{ah}{2} }$

    Applying formula   $\lim_{h \to \ 0} \frac{sin ( \frac{ah}{2} )}{\frac{ah}{2} } =1$

     $= -2 sin( ax+\frac{a(0)}{2} ) .\frac{a}{2} (1)$

    $= -a sin(ax+0)$

    $=-a sin\ ax$

Thus, cos ax derivative by first principle method is  - a sin ax.