Question

Find the second order derivative of y=log(ax+x^2)

Answer 1

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Answer 2

Answer:

here is your answer

Step-by-step explanation:

Let y=log(logx)

⇒

dx

dy

=

dx

d

[log(logx)]=

logx

1

.

dx

d

(logx)=

xlogx

1

=(xlogx)

−1

∴

dx

2

d

2

y

=

dx

d

[(xlogx)

−1

]=(−1).(xlogx)

−2

.

dx

d

(xlogx)

=

(xlogx)

2

−1

.[logx.

dx

d

(x)+x

dx

d

(logx)]

=

(xlogx)

2

−1

[logx.1+x

x

1

]=

(xlogx)

2

−(1+logx)

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