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How do I find the derivative of #y= x^(2x+1)#?

Answer 1

Emily Ammerman

The rule to derivate exponential functions is

If #y=a^u#, then #y'=a^u(lna)u'#

Solution:

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

#(dy)/(dx)=2x^(2x+1)*lnx#

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Answer 2

Isabella Ackerman

To find the derivative of ( y = x^{2x+1} ), you will need to use logarithmic differentiation.

[ y = x^{2x+1} ]

Take the natural logarithm of both sides:

[ \ln(y) = \ln(x^{2x+1}) ]

Apply the power rule of logarithms:

[ \ln(y) = (2x+1) \ln(x) ]

Differentiate implicitly with respect to ( x ):

[ \frac{1}{y} \frac{dy}{dx} = 2 \ln(x) + (2x+1) \frac{1}{x} ]

Substitute back ( y = x^{2x+1} ):

[ \frac{dy}{dx} = y \left(2 \ln(x) + \frac{2x+1}{x}\right) ]

Replace ( y ) with ( x^{2x+1} ):

[ \frac{dy}{dx} = x^{2x+1} \left(2 \ln(x) + \frac{2x+1}{x}\right) ]

This is the derivative of ( y = x^{2x+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.