How do you differentiate #g(x)=cos(10^(2x))#?

1 Answer
Jul 25, 2017

#g'(x)=-2ln10sin(10^(2x))10^(2x)#

Explanation:

#•color(white)(x)d/dx(a^x)=a^xlna#

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(h(x)))" then"#

#dy/dx=f'(g(h(x))xxg'(h(x))xxh'(x)larr" chain rule"#

#rArrg'(x)=-sin(10^(2x))xxd/dx(10^(2x))#

#color(white)(rArrg'(x))=-sin(10^(2x))xx(10^(2x))ln10xxd/dx(2x)#

#color(white)(rArrg'(x))=-2ln10sin(10^(2x))10^(2x)#