How do you differentiate #5^(2x^2)#?

Answer 1

Christopher Allers

#d/dx (5^(2x^2)) = 4ln5 x 5^(2x^2)#

Consider that:

#5^(2x^2) = (e^ln5)^(2x^2) = e^(2ln5x^2)#

So, using the chain rule:

#d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = e^(2ln5x^2) d/dx (2ln5x^2)#

#d/dx (5^(2x^2)) = d/dx(e^(2ln5x^2)) = 4ln5xe^(2ln5x^2) = 4ln5 x 5^(2x^2)#

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Answer 2

Isaiah Allen

#(dy)/(dx)=5^(2x^2)4xln5#

#y=5^(2x^2)#

take natural logs of both sides

#lny=ln5^(2x^2)#

using laws of logs

#lny=2x^2ln5#

differentiate implicitly

#1/y(dy)/(dx)=4xln5#

#(dy)/(dx)=y4xln5#

#:.(dy)/(dx)=5^(2x^2)4xln5#

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Answer 3

Ezekiel Alexander

To differentiate ( 5^{2x^2} ), you would use the chain rule. First, take the natural logarithm of the expression, then differentiate with respect to ( x ), and finally, multiply the result by the derivative of the exponent ( 2x^2 ). The derivative is:

[ \frac{d}{dx} \left(5^{2x^2}\right) = \left(5^{2x^2}\right) \cdot \left(2 \cdot 2x \cdot \ln(5)\right) = 4x \ln(5) \cdot 5^{2x^2} ]

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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.