Question

Differentiable x^x with respect to x^ sinx​

Answer 1

Answer:

Let y=x

sinx

Apply logarithms on both sides

logy=sinx⋅logx

Diff w.r.t x

y

1

dx

dy

=sinx⋅

dx

d

(logx)+logx⋅

dx

d

(sinx)

y

1

dx

dy

=sinx⋅

x

1

+logx⋅cosx

dx

dy

=y[

x

sinx

+logx⋅cosx]

∴

dx

dy

=x

sinx

[

x

sinx

+logx⋅cosx]

Answer 2

Answer:

Let y=x

sinx

Step-by-step explanation:

Apply logarithms on both sides

logy=sinx⋅logx

Diff w.r.t x

y

1

dx

dy

=sinx⋅

dx

d

(logx)+logx⋅

dx

d

(sinx)

y

1

dx

dy

=sinx⋅

x

1

+logx⋅cosx

dx

dy

=y[

x

sinx

+logx⋅cosx]

∴

dx

dy

=x

sinx

[

x

sinx

+logx⋅cosx].