Question

if y=(x)^x + (sinx)^x then find dy/dx​

Answer 1

Answer:

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Answer 2

Answer:

Given y=(sinx)

x

+x

x

Let u=x

x

and v=(sinx)

x

Therefore y=u+v

Differentiating above equation,

dx

dy

=

dx

du

+

dx

dv

Since u=x

x

Taking log on both sides, logu=xlogx

Differentiating both,

u

1

dx

dy

=x

x

1

+logx

dx

dy

=u(1+logx)=x

x

(1+logx)

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