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

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

Applying the product rule we get:

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To find the derivative of ( y = x^n \ln x ), you can use the product rule. The product rule states that if ( u ) and ( v ) are both functions of ( x ), then the derivative of their product is given by ( (uv)' = u'v + uv' ). Applying this rule to the function ( y = x^n \ln x ), we get:

[ y' = (x^n)' \ln x + x^n (\ln x)' ]

Now, differentiate each term separately using the power rule and the derivative of natural logarithm, which is ( (\ln x)' = \frac{1}{x} ):

[ y' = (n x^{n-1}) \ln x + x^n \cdot \frac{1}{x} ]

[ y' = n x^{n-1} \ln x + \frac{x^n}{x} ]

[ y' = n x^{n-1} \ln x + x^{n-1} ]

Therefore, the derivative of ( y = x^n \ln x ) with respect to ( x ) is ( y' = n x^{n-1} \ln x + x^{n-1} ).

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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.

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.

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.

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