# If # y = (sinx)^(sinx) # then find #dy/dx#?

##### 4 Answers

#### Explanation:

We want to find the derivative of

Take the logarithm on both sides

Differentiate both sides using the product and chain rule

(Be aware of the implicit differentiation on the right side)

Substitute

#### Explanation:

#### Explanation:

We have to differentiate

Or maybe,

Remember,

So the above becomes

Apply the chain rule:

The rule states that

Here,

The equation simplifies (or gets excruciatingly confusing) to

The derivative of

Now for the second half. Remember the product rule:

So now the equation is:

The derivative of

So the first derivative is

Sadly, we must use the chain rule again. Here, I take it as the differentiation of

The derivative of

We now have

So this becomes

Our equation is presently

But

We now have

The equation is

But

So now, the equation is

This, (thank God), cannot be simplified further.

# dy/dx = cosx \ (sinx)^(sinx) \ {1+ ln sinx }#

#### Explanation:

We have:

# y = (sinx)^(sinx) #

If we take Natural Logarithms, we have:

# ln y = ln {(sinx)^(sinx) }#

And using the properties of logarithms we have:

# ln y = sinx \ ln sinx#

We can now readily differentiate wrt

# 1/y \ dy/dx = (sinx)(1/sinx cosx) + (cosx)ln sinx #

Which we can simplify:

# 1/y \ dy/dx = cosx + cosx \ ln sinx #

# :. dy/dx = y{cosx + cosx \ ln sinx }#

# \ \ \ \ \ \ \ \ \ \ = ycosx{1+ ln sinx }#

# \ \ \ \ \ \ \ \ \ \ = cosx \ (sinx)^(sinx) \ {1+ ln sinx }#