please solve this differentiation

If f(x)=a^x ,

then f^'(x) = a^xloga

So,

for f(x)=3^(x+2)

f^'(x) = 3^(x+2)log3

please solve this one also

y =

 f(x)=a^x ,

=>f^'(x) = a^xloga

=> f(x)=3^(x+2)

=>f^'(x) = 3^(x+2)log3

y =

take logarithm on both sides with base e

log y= log (e ^xcos x)

log y = x *cos x

differentiating

(1/y )dy/dx = -x sin x +cosx

=> dy/dx =  ( -x sin x +cosx)

ok i will tell u the general principle...

tak ln on both sides u get lny = xcosx now diff.

(1/y)y' = -xsinx+cosx => y' = e^xcosx(cosx-xsinx)....u can use it for ur previous q and any where u have a^b

y =

hey same

take logon both sides

log y= log (e ^xcos x)

log y = x *cos x

differentiating

(1/y )dy/dx = -x sin x +cosx

=> dy/dx = ( -x sin x +cosx)

y = e^xcosx

Differentiating both sides and by using the chain rule,we have........