Question

Differentiate this with respect to X


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Answer 1

ANSWER

y' = d/dx(sin^x(x)+(x cos(x))^x)

Differentiate the sum term by term:

y'(x) = d/dx(sin^x(x))+d/dx((x cos(x))^x)

Use the chain rule,

d/dx((x cos(x))^x) = ( du^v)/( du) ( du)/( dx)+( du^v)/( dv) ( dv)/( dx),

where u = x cos(x), v = x and ( du^v)\/( du) = u^(-1+v) v, ( du^v)\/( dv) = u^v log(u):

y' = d/dx(sin^x(x))+ (x (x cos(x))^(x-1) (d/dx(x cos(x)))+(x cos(x))^x log(x cos(x)) (d/dx(x)))

Use the chain rule,

y' = x (x cos(x))^(x-1) (d/dx(x cos(x))) + (x sin^(x-1)(x) (d/dx(sin(x)))+sin^x(x) log(sin(x)) (d/dx(x)))+(x cos(x))^x log(x cos(x)) (d/dx(x))

Use the product rule,

y' = x sin^(x-1)(x) (d/dx(sin(x))) + x (x cos(x))^(x-1) (x (d/dx(cos(x))) + cos(x) (d/dx(x)))+sin^x(x) log(sin(x)) (d/dx(x)) + (x cos(x))^x log(x cos(x)) (d/dx(x))

y' = x sin^(x-1)(x) (d/dx(sin(x))) +x (x cos(x))^(x-1) (x (d/dx(cos(x)))+cos(x))+sin^x(x) log(sin(x)) (d/dx(x))+(x cos(x))^x log(x cos(x)) (d/dx(x))

y' = x sin^(x-1)(x) (d/dx(sin(x))) + sin^x(x) log(sin(x)) (d/dx(x)) + (x cos(x))^x log(x cos(x)) (d/dx(x))+x (x cos(x))^(x-1) (x (-sin(x))+cos(x))

y' = x sin^(x-1)(x) (d/dx(sin(x)))+sin^x(x) log(sin(x)) (d/dx(x))+(x cos(x))^x log(x cos(x))+x (x cos(x))^(x-1) (cos(x)-x sin(x))

y' = sin^x(x) log(sin(x)) (d/dx(x))+(x cos(x))^x log(x cos(x))+x sin^(x-1)(x) cos(x)+x (x cos(x))^(x-1) (cos(x)-x sin(x))

y' = sin^x(x) log(sin(x))+(x cos(x))^x log(x cos(x))+x sin^(x-1)(x) cos(x)+x (x cos(x))^(x-1) (cos(x)-x sin(x)).....................answer

Answer 2

Explanation:

Check out the given attachment