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The power rule for differentiation was established for positive integers and it was discussed that it could be proved, one way, for negative integers by using rules of exponents and logarithms. However, there is another way! Use the derivative quotient rule to prove the power rule for negative integers, that is,

d/dx (x⁻ᵐ)=-mx⁻ᵐ⁻¹ Where m is a positive integer. 15. Is there a value of c that will make f(x) continuous at x=0 ? f(x)={sin²3x/x², x ≠ 0
{c, x = 0

User Boris Smus
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Final answer:

The power rule for differentiation can be proved using the derivative quotient rule for negative integers. To make the function continuous at x=0, the value of c should be equal to the limit of the function as x approaches 0, which is 0.

Step-by-step explanation:

To prove the power rule for negative integers using the derivative quotient rule, we need to apply the quotient rule to the function f(x) = 1/x^m, where m is a positive integer. Using the quotient rule, we get:



f'(x) = (x^m * 0 - 1 * (m * x^(m-1))) / (x^m)^2 = -mx^(m-1) / x^2 = -mx^(m-1-2) = -mx^(-m-1).



Therefore, d/dx(x^(-m)) = -mx^(-m-1), proving the power rule for negative integers.



To find a value of c that will make f(x) continuous at x=0, we need to find the limit of f(x) as x approaches 0. By evaluating the limit, we can determine the value of c. Since the numerator, sin^2(3x), approaches 0 as x approaches 0, to make f(x) continuous, the value of c should be equal to that limit, which is 0. Therefore, f(x) = 0 for x=0 to make it continuous.

Learn more about power rule for negative integers

User Jack Jia
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