y= 7x* (cosx)^x/2
To solve we
will apply the natural logarithm to both sides.
==>
ln y= ln (7x*(cosx)^x/2)
Now we know that ln a*b = ln a +
ln b
==> ln y= ln 7x + ln
(cosx)^x/2
Now we know that ln x^a = aln
x
==> ln y= ln 7x + x/2 *ln
cosx
Now we will differentiate both
sides.
==> 1/y *y' = 7*1/7x + (x/2)'*ln cosx +
(x/2)*(ln cosx)'
==> 1/y y' = 1/x + (1/2)*ln cosx +
(x/2)* -sinx / cosx
==> 1/y y' = 1/x + ln cosx /2 -
xtanx /2
==> 1/y y' = ( 2+ xln cosx -
x^2*tanx)/2x
Now we will multiply by y=
7x*cosx)^x/2
==> y' = (2+ xln cosx - x^2 *tanx)/2x
* 7x*cosx^x/2
==> y' =(7/2)* ( 2+ xln
cosx - x^2 tanx) *(cosx)^x/2
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