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Grade: 12
        
can anyone please tell me what will be the
dy/dx of x^(sinx)+ cosx^(tanx)
one month ago

Answers : (1)

Aditya Gupta
1670 Points
							
whenever we have to differentiate f(x)^g(x), we can use the following method:
let z= f(x)^g(x)
take log
ln z= g(x)*ln f(x)
differentiate using chain rule
(1/z)*dz/dx= g(x)*ln f(x) + g(x)*f(x)/f(x)
so that dz/dx= f(x)^g(x)[g(x)*ln f(x) + g(x)*f(x)/f(x)]
so now simply see that dy/dx= d(x^(sinx))/dx + d(cosx^(tanx))/dx
these both can be easily differentiated since we know how individually differentiate each of x, sinx, cosx and tanx.
kindly approve :)
one month ago
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