Question

differentiate x^x+ (sinx)^lnx

Answer 1

Answer:

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

Step-by-step explanation:

The trick to handling situations where the variable occurs both in the base of the power as well as the exponent is to take logs, so we get a product, and then use the product rule.

Deal with the terms one at a time.

First the x^x.  Let's write y = x^x.  Taking logs, this becomes

$\ln y = x \ln x$

Now we differentiatie (chain rule on the left and product rule on the right) to get

$\frac{1}{y}\frac{dy}{dx} = x\frac{1}{x} + \ln x = 1 + \ln x \\\\\Rightarrow \frac{dy}{dx} = y(1+\ln x) = x^x(1+\ln x)$

Now the second term.  Again, let's put y = (sin x)^(ln x).  Taking logs, we get

$\ln y = \ln x \ln(\sin x)$

Preparing for what's about to come, let's sort out the derivative of ln(sin(x)).  Using the chain rule, this will be ( 1 / sin(x) ) × cos(x) = cot(x).  Great, so now we can differentiate the equation above, using the product rule on the right and the chain rule on the left just like last time:

$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\ln(\sin x) + \ln x \cot x \\\\\Rightarrow \frac{dy}{dx} = \left(\frac{1}{x}\ln(\sin x) + \ln x \cot x\right) (\sin x)^{\ln x}$

Adding the derivatives of the two terms gives the required result.