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How do you differentiate # x^sin(x)#?

Answer 1

Luke Alvarez

#(dy)/(dx)=(x^sinx)(cosxlnx+sinx/x)#

let #y=x^sinx#

take natural logarithms to both sides and simplify

#lny=lnx^sinx#

#=>lny=sinxlnx#

differentiate both sides wrt #x#

#d/(dx)(lny)=d/(dx)(sinxlnx)#

using implicit differentiation on the LHS; product rule on RHS

#=1/y(dy)/dx=cosxlnx+sinx/x#

#=>(dy)/(dx)=y(cosxlnx+sinx/x)#

substituting back for #y#

#(dy)/(dx)=(x^sinx)(cosxlnx+sinx/x)#

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Answer 2

Ezekiel Alexander

To differentiate (x^{\sin(x)}), you would use the chain rule. The chain rule states that if you have a function (f(g(x))), then its derivative is (f'(g(x)) \cdot g'(x)).

So, for (x^{\sin(x)}), let's denote (f(u) = u^{\sin(x)}) and (g(x) = x). Then (f'(u) = \sin(x) \cdot u^{\sin(x)-1}) and (g'(x) = 1).

Applying the chain rule, the derivative of (x^{\sin(x)}) is:

[ \frac{d}{dx}\left(x^{\sin(x)}\right) = \sin(x) \cdot x^{\sin(x)-1} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.