Finding Derivative Functions and Values

Question

Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.

g(t) = 1/t²; g′(−1), g′(2), g′(√3)

Accepted Answer

Welcome back everyone. Given the function G of X equals 18 divided by X2, find G of X using the limit definition of the derivative. Then calculate the value of G 5. Let's refold the definition of the derivative. G of X can be calculated as limit as H approaches 0 of G at X plus H minus G of X divided by H. We're going to use the given function, so we get the limit as H approaches 0 of G X plus H, which is just 18 divided by X + H2, and then we are subtracting the original function 18 divided by X2. All of that is divided by H. Let's rewrite the limit as H approaches 0. And for our numerator, we're going to use the common denominator, which is X2 multiplied by X plus H2, and we're going to combine the numerators. So the first one becomes 18 multiplied by X2. And the second one becomes negative, 18. Multiplied by X plus H2. And all of that is divided by H. To simplify because we have a double division, we can just move that age into the denominator of X2 + X2 multiplied by X + a2, right? Since we have a double division, we can just combine those denominators into one. Now from here, we're going to factor out 18 because it's a constant, we can see that, so we got 18. Multiplied by a limit as H approaches 0. Of x2 minus X + H2. Divided by H X squared multiplied by X plus H2. Let's apply the difference of squares formula and the. Numerator we can factor it out, right? So we got 18 limit as H approaches 0 of X plus X plus H multiplied by X minus X minus H. And this is divided by HX2 multiplied by X plus. H square. We can notice that X minus X gives us 0, and then X plus X gives us 2 X. So we can rewrite our limit as 18 limit as H approaches 0 off. 2 X plus H multiplied by negative H. And we are dividing by HX2 multiplied by X plus H2. So from here we can cancel out the H term, right? Because it's a common factor within our fraction. We get 18. Multiplied by limit as H approaches 0 of negative 2 X plus H. That's our numerator and our denominator is X2 multiplied by X plus H2. And we know That each approach is 0, so we're going to perform direct substitution which gives us 18. Multiplied by -2 X + 0, so just 2 X divided by X2 multiplied by X + 02. So that's X2 multiplied by X2, which is X rated to the 4th power. Now multiplying, we're going to get negative 36 divided by X cube. Well then, so we have G of X, and now we are asked to identify G of 5, which means that we're substituting X equals 5. Let's substitute, this gives us 36 divided by 5 cubed. Which is equal to. Negative. 36 divided by 125. Well then, we have arrived at the final answers. We have our derivative -36 divided by X cubed and the value of the derivative at 5, which is 36 divided by 125. Thank you for watching.

Finding Derivative Functions and Values

Using the definition, calculate the derivatives of the functions in Exercises 1–6. Then find the values of the derivatives as specified.

g(t) = 1/t²; g′(−1), g′(2), g′(√3)

Key Concepts

Definition of Derivative

Power Rule for Derivatives

Substitution in Derivatives

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