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How do you differentiate #f(x) = sin ( x² ln(x) )#?

Answer 1

Elijah Abram

#(df(x))/dx = x(2lnx+1)cos(x^2lnx)#

#(df(x))/dx = d/dx(sin(x^2lnx))=cos(x^2lnx)*(2xlnx+x^2 1/x)#

#(df(x))/dx = x(2lnx+1)cos(x^2lnx)#

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Answer 2

Luke Alvarez

To differentiate ( f(x) = \sin(x^2 \ln(x)) ), you would use the chain rule along with the product rule. Here's the process:

Apply the chain rule to differentiate the outer function, which is ( \sin(u) ), where ( u = x^2 \ln(x) ).
The derivative of ( \sin(u) ) with respect to ( u ) is ( \cos(u) ).
Then, differentiate the inner function ( u = x^2 \ln(x) ) using the product rule, where ( f(x) = x^2 ) and ( g(x) = \ln(x) ).
The derivative of ( f(x) = x^2 ) is ( f'(x) = 2x ), and the derivative of ( g(x) = \ln(x) ) is ( g'(x) = \frac{1}{x} ).
Apply the product rule: ( (f \cdot g)' = f'g + fg' ).
Combine the results using the chain rule and product rule.

The derivative of ( f(x) ) is:

[ f'(x) = \cos(x^2 \ln(x)) \cdot (2x \ln(x) + x) ]

[ + \sin(x^2 \ln(x)) \cdot \left(2 + \frac{2x}{x}\right) ]

[ = \cos(x^2 \ln(x)) \cdot (2x \ln(x) + x) + 2x \sin(x^2 \ln(x)) ]

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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.