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How do you differentiate #y = 10^(1-x^2)#?

Answer 1

Ezra Adams

# y' = -2x * ln(10) * 10^(1-x^2)#

#y = 10^(1-x^2)#
#ln y = (1-x^2)ln(10)#
#1/y \ y' = -2x * ln(10)#
# y' = -2x * ln(10) * 10^(1-x^2)#

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Answer 2

Ezra Adams

To differentiate ( y = 10^{1-x^2} ), you can use the chain rule. First, take the natural logarithm of both sides to simplify the expression. Then, apply the chain rule to find the derivative of the function with respect to ( x ). The chain rule states that if ( y = f(g(x)) ), then ( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) ).

So, the steps are as follows:

Take the natural logarithm of both sides: [ \ln(y) = \ln(10^{1-x^2}) = (1-x^2) \ln(10) ]
Differentiate both sides with respect to ( x ): [ \frac{1}{y} \frac{dy}{dx} = -2x \ln(10) ]
Multiply both sides by ( y ): [ \frac{dy}{dx} = -2xy \ln(10) ]
Substitute the original expression for ( y ): [ \frac{dy}{dx} = -2x \cdot 10^{1-x^2} \ln(10) ]

So, the derivative of ( y = 10^{1-x^2} ) with respect to ( x ) is ( -2x \cdot 10^{1-x^2} \ln(10) ).

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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.