# What is the derivative of y= 7x*(cosx)^x/2

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### 1 Answer

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**