Question

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

Answer 1

`y'=sin(x)^sin(x)cos(x)*(ln(sin(x))+1)`

Explanation:

We want to find the derivative of

`y=(sin(x))^sin(x)`

Take the logarithm on both sides

`ln(y)=ln((sin(x))^sin(x))`

`ln(y)=sin(x)*ln(sin(x))`

Differentiate both sides using the product and chain rule

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

`y'*1/y=cos(x)ln(sin(x))+sin(x)cos(x)*1/sin(x)`

`y'=y*(cos(x)ln(sin(x))+sin(x)cos(x)*1/sin(x))`

`y'=y*(cos(x)ln(sin(x))+cos(x))`

`y'=y*cos(x)*(ln(sin(x))+1)`

Substitute `y=(sin(x))^sin(x)`

`y'=(sin(x))^sin(x)cos(x)*(ln(sin(x))+1)`

Answer 2

`dy/dx= (sinx)^sinx (cosx+ cosx log sinx)`

Explanation:

`y= (sinx)^sinx` Taking log on both sides we get ,

`log y = sinx log sinx` Differentiating both sides we get ,

`1/y*dy/dx= sinx * 1/sinx *cosx+ cosx log sinx`or

`1/y*dy/dx= cancel(sinx) * 1/cancel(sinx) *cosx+ cosx log sinx` or

`dy/dx= y (cosx+ cosx log sinx)` or

`dy/dx= (sinx)^sinx (cosx+ cosx log sinx)` [Ans]

Answer 3

`sin^sin(x)(x)(cos(x)lnsin(x)cos(x))`

Explanation:

We have to differentiate `sin(x)^sin(x)`.

Or maybe, `d/dxsin^sin(x)(x)`.

Remember, `a^b=e^(b*lna)`.

So the above becomes `d/dxe^(sin(x)*lnsin(x)`

Apply the chain rule:

The rule states that `(df(u))/dx=(df)/(du)*(du)/dx`.

Here, `f=e^u` and `u=sin(x)*lnsin(x)`

The equation simplifies (or gets excruciatingly confusing) to

`d/(du)e^u*d/dxsin(x)*lnsin(x)`

The derivative of `e^x` is `e^x`. So now the equation is:

`e^u*d/dxsin(x)*lnsin(x)`

Now for the second half. Remember the product rule: `(f*g)'=f'g+fg'`.

So now the equation is:

`e^u*d/dx(sin(x))*lnsin(x)+d/dx(lnsin(x))*sin(x)`

The derivative of `sin(x)` is `cos(x)`.

So the first derivative is `cos(x)lnsin(x)`. The equation is now `e^ucos(x)lnsin(x)*d/dx(lnsin(x))*sin(x)`

Sadly, we must use the chain rule again. Here, I take it as the differentiation of `f(w)`.

`f=lnw`, and `w=sin(x)`

The derivative of `lnw` is `1/w`, and `sin(x)` is again `cos(x)`

We now have `cos(x)/w`. But remember, `w=sin(x)`.

So this becomes `cos(x)/sin(x)`, which simplifies to `cot(x)`.

Our equation is presently `e^ucos(x)lnsin(x)cot(x)sin(x)`.

But `cot(x)sin(x)=cos(x)/sin(x)*sin(x)=cos(x)`

We now have `e^ucos(x)lnsin(x)cos(x)`. We're not done yet, though.

`u=sin(x)lnsin(x)`.

The equation is `e^(sin(x)lnsin(x))cos(x)lnsin(x)cos(x)`.

But `e^(b*lna)=a^b`.

So now, the equation is `sin^sin(x)(x)(cos(x)lnsin(x)cos(x))`

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

Answer 4

`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 `x` by applying the chain rule (or implicit differentiation the LHS and the chain rule and the product rule on the RHS:

`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 }`