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

##### 4 Answers
Jan 30, 2018

$y ' = \sin {\left(x\right)}^{\sin} \left(x\right) \cos \left(x\right) \cdot \left(\ln \left(\sin \left(x\right)\right) + 1\right)$

#### Explanation:

We want to find the derivative of

$y = {\left(\sin \left(x\right)\right)}^{\sin} \left(x\right)$

Take the logarithm on both sides

$\ln \left(y\right) = \ln \left({\left(\sin \left(x\right)\right)}^{\sin} \left(x\right)\right)$

$\ln \left(y\right) = \sin \left(x\right) \cdot \ln \left(\sin \left(x\right)\right)$

Differentiate both sides using the product and chain rule

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

$y ' \cdot \frac{1}{y} = \cos \left(x\right) \ln \left(\sin \left(x\right)\right) + \sin \left(x\right) \cos \left(x\right) \cdot \frac{1}{\sin} \left(x\right)$

$y ' = y \cdot \left(\cos \left(x\right) \ln \left(\sin \left(x\right)\right) + \sin \left(x\right) \cos \left(x\right) \cdot \frac{1}{\sin} \left(x\right)\right)$

$y ' = y \cdot \left(\cos \left(x\right) \ln \left(\sin \left(x\right)\right) + \cos \left(x\right)\right)$

$y ' = y \cdot \cos \left(x\right) \cdot \left(\ln \left(\sin \left(x\right)\right) + 1\right)$

Substitute $y = {\left(\sin \left(x\right)\right)}^{\sin} \left(x\right)$

$y ' = {\left(\sin \left(x\right)\right)}^{\sin} \left(x\right) \cos \left(x\right) \cdot \left(\ln \left(\sin \left(x\right)\right) + 1\right)$

Jan 30, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{\sin} x \left(\cos x + \cos x \log \sin x\right)$

#### Explanation:

$y = {\left(\sin x\right)}^{\sin} x$ Taking log on both sides we get ,

$\log y = \sin x \log \sin x$ Differentiating both sides we get ,

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sin x \cdot \frac{1}{\sin} x \cdot \cos x + \cos x \log \sin x$or

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \cancel{\sin x} \cdot \frac{1}{\cancel{\sin x}} \cdot \cos x + \cos x \log \sin x$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(\cos x + \cos x \log \sin x\right)$ or

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{\sin} x \left(\cos x + \cos x \log \sin x\right)$ [Ans]

Jan 30, 2018

${\sin}^{\sin} \left(x\right) \left(x\right) \left(\cos \left(x\right) \ln \sin \left(x\right) \cos \left(x\right)\right)$

#### Explanation:

We have to differentiate $\sin {\left(x\right)}^{\sin} \left(x\right)$.

Or maybe, $\frac{d}{\mathrm{dx}} {\sin}^{\sin} \left(x\right) \left(x\right)$.

Remember, ${a}^{b} = {e}^{b \cdot \ln a}$.

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

Apply the chain rule:

The rule states that $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Here, $f = {e}^{u}$ and $u = \sin \left(x\right) \cdot \ln \sin \left(x\right)$

The equation simplifies (or gets excruciatingly confusing) to

$\frac{d}{\mathrm{du}} {e}^{u} \cdot \frac{d}{\mathrm{dx}} \sin \left(x\right) \cdot \ln \sin \left(x\right)$

The derivative of ${e}^{x}$ is ${e}^{x}$. So now the equation is:

${e}^{u} \cdot \frac{d}{\mathrm{dx}} \sin \left(x\right) \cdot \ln \sin \left(x\right)$

Now for the second half. Remember the product rule: $\left(f \cdot g\right) ' = f ' g + f g '$.

So now the equation is:

${e}^{u} \cdot \frac{d}{\mathrm{dx}} \left(\sin \left(x\right)\right) \cdot \ln \sin \left(x\right) + \frac{d}{\mathrm{dx}} \left(\ln \sin \left(x\right)\right) \cdot \sin \left(x\right)$

The derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$.

So the first derivative is $\cos \left(x\right) \ln \sin \left(x\right)$. The equation is now ${e}^{u} \cos \left(x\right) \ln \sin \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(\ln \sin \left(x\right)\right) \cdot \sin \left(x\right)$

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

$f = \ln w$, and $w = \sin \left(x\right)$

The derivative of $\ln w$ is $\frac{1}{w}$, and $\sin \left(x\right)$ is again $\cos \left(x\right)$

We now have $\cos \frac{x}{w}$. But remember, $w = \sin \left(x\right)$.

So this becomes $\cos \frac{x}{\sin} \left(x\right)$, which simplifies to $\cot \left(x\right)$.

Our equation is presently ${e}^{u} \cos \left(x\right) \ln \sin \left(x\right) \cot \left(x\right) \sin \left(x\right)$.

But $\cot \left(x\right) \sin \left(x\right) = \cos \frac{x}{\sin} \left(x\right) \cdot \sin \left(x\right) = \cos \left(x\right)$

We now have ${e}^{u} \cos \left(x\right) \ln \sin \left(x\right) \cos \left(x\right)$. We're not done yet, though.

$u = \sin \left(x\right) \ln \sin \left(x\right)$.

The equation is ${e}^{\sin \left(x\right) \ln \sin \left(x\right)} \cos \left(x\right) \ln \sin \left(x\right) \cos \left(x\right)$.

But ${e}^{b \cdot \ln a} = {a}^{b}$.

So now, the equation is ${\sin}^{\sin} \left(x\right) \left(x\right) \left(\cos \left(x\right) \ln \sin \left(x\right) \cos \left(x\right)\right)$

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

Jan 30, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \setminus {\left(\sin x\right)}^{\sin x} \setminus \left\{1 + \ln \sin x\right\}$

#### Explanation:

We have:

$y = {\left(\sin x\right)}^{\sin x}$

If we take Natural Logarithms, we have:

$\ln y = \ln \left\{{\left(\sin x\right)}^{\sin x}\right\}$

And using the properties of logarithms we have:

$\ln y = \sin x \setminus \ln \sin x$

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:

$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\sin x\right) \left(\frac{1}{\sin} x \cos x\right) + \left(\cos x\right) \ln \sin x$

Which we can simplify:

$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x + \cos x \setminus \ln \sin x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\cos x + \cos x \setminus \ln \sin x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = y \cos x \left\{1 + \ln \sin x\right\}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \cos x \setminus {\left(\sin x\right)}^{\sin x} \setminus \left\{1 + \ln \sin x\right\}$