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How do you find the derivative of #y=x^2+x^x#?

Answer 1

We know the first half. Let's try the second half, using a more foolproof way than simply trying to remember the derivative of a similar function.

Let: #y = x^x#

#log_x(y) = log_x(x^(x)) = x#

#(lny)/(lnx) = x#

#lny = xlnx#

#1/y ((dy)/(dx)) = x*1/x + lnx*1# (Implicit Differentiation, Product Rule)

#1/y ((dy)/(dx)) = 1 + lnx#

#((dy)/(dx))_(y = x^x) = y[1 + lnx]#

#= x^x[1 + lnx]#

Thus you have overall:

#color(blue)(((dy)/(dx))_(y = x^2 + x^x) = 2x + x^x(1+lnx))#

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Answer 2

Ezra Adams

To find the derivative of ( y = x^2 + x^x ), you would use the sum rule and the chain rule.

The derivative of ( x^2 ) is ( 2x ).
The derivative of ( x^x ) requires the chain rule. Let ( u = x ) and ( v = x ), then ( x^x = u^v ). The derivative of ( u^v ) is ( v \cdot u^{v-1} \cdot u' + \ln(u) \cdot u^v \cdot v' ), where ( u' ) and ( v' ) are the derivatives of ( u ) and ( v ), respectively. Applying this to ( x^x ), we get ( x^x (\ln(x) + 1) ).

Therefore, the derivative of ( y = x^2 + x^x ) is ( 2x + x^x (\ln(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.