How to differentiate x^sinx ?

Answer 1

Logarithmic differentiation.

y = x^{sin x}

ln y = (sin x) ln x     ←     taking the natural log of both sides.

y'/y = (cos x) ln x + (sin x)/x     ←     Using the product rule and implicit differentiation.

y' = y((cos x) ln x + (sin x)/x)     ←     multiplying by y.

y' = x^{sin x} ((cos x) ln x + (sin x)/x) ← substitution.

Note that if you use logarithmic differentiation to differentiate y = u^v with respect to x, you will get the following:

y' = vu^v-1 * du/dx + u^v ln u * dv/dx.

Notice anything familiar?

Answer 2

x^sin(x)[cos(x)ln(x) + sin(x)/x].

 

You can write the function y =x^sin(x) as  

y = exp(log(x^sin(x)))= e^sinx*log(x),

then dy/dx = [sin(x)/x + ln(x)cos(x)]e^sin(x)log(x) = [sin(x)/x + ln(x)cos(x)]x^sin(x)  .

Answer 3

Hello Mike,

Often, to differentiate a function y = f(x) with a variable base and exponent, you should make use of the identity

g(x)^h(x) = e^{h(x) ln g(x)}

and use your know rules for differentiating the exponential function to proceed.  You will need to use the Product Rule for Differentiation to differentiate the exponent on the right-hand side of the identity above.

Another (equivalent) way is to "take the logarithm of both sides"; namely, for y = g(x)^h(x), write

ln y = h(x) ln g(x)

and then differentiate using your known rule for the logarithm.  Note that you will obtain

y'/y

as the derivative on the left, for which you'll then solve for y' and substitute the original expression for y back in.  Good luck!

Regards,

Hassan H.