How do you differentiate #y=2^(sinpix)#?
3 Answers
Explanation:
You must write
Now we differentiate :
Explanation:
use lograthmic differentiation
differentiate implicitly
substitue back for
tidying up
Explanation:
#•color(white)(x)d/dx(a^x)=a^xlna#
#•color(white)(x)d/dx(a^(f(x)))=a^(f(x))lnaxxf'(x)larr" chain rule"#
#y=2^(sinpix)#
#rArrdy/dx=2^(sinpix)ln2xxd/dx(sinpix)#
#color(white)(rArrdy/dx)=2^(sinpix)ln2xxcos(pix)xxd/dx(pix)#
#color(white)(rArrdy/dx)=piln2cos(pix)2^(sinpix)#