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How do you differentiate #y=x^(8x)#?

Answer 1

Charles Anctil

Use logarithmic differentiation (or its equivalent exponential form).

#y=x^(8x)#

Logarithmic Differentiation

Take the logarithm of both sides.

#lny=ln(x^(8x))=8xlnx#. Now differentiate implicitly.

#1/y dy/dx = 8lnx+8x(1/x) = 8lnx+8#

#dy/dx = y(8lnx+8) = x^(8x)(8lnx+8)#.

Exponential Equivalent

#y=x^(8x) = e^(ln(x^(8x))) = e^(8xlnx)#

Differentiate using #d/dx(e^u) = e^u(du)/dx#

#dy/dx = e^(8xlnx)(8lnx+8)#

# = x^(8x)(8lnx+8)#

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Answer 2

Isaiah Allen

To differentiate ( y = x^{8x} ), you would use logarithmic differentiation. First, take the natural logarithm of both sides: ( \ln y = \ln(x^{8x}) ). Then apply the properties of logarithms to simplify: ( \ln y = 8x \ln x ). Differentiate both sides with respect to ( x ): ( \frac{1}{y} \frac{dy}{dx} = 8 \ln x + 8 ). Solve for ( \frac{dy}{dx} ): ( \frac{dy}{dx} = y(8 \ln x + 8) ). Substitute ( y = x^{8x} ): ( \frac{dy}{dx} = x^{8x}(8 \ln x + 8) ).

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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.