. Here f(x) = −10x^2 and f(x + h) = −10(x + h)^2 = −10(x^2 + 2xh + h^2 ) =
−10x^2 − 20xh − 10h^2.If we now plug these into the definition of the derivative, we get f′(x) = = . This simplifies to f′(x) = . Now wecan factor out the h from the numerator and cancel it with the h in the denominator: f′(x) = (−20x − 10h). Now we take the limit to get f′(x) = (−20x − 10h)= −20x.- 40 a
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = 20x^2 and x = a. This means that f(a) = 20a^2 and f(a + h) =
20(a + h)^2 = 20(a^2 + 2ah + h^2 ) = 20a^2 + 40ah + 20h^2 . If we now plug these into the definitionof the derivative, we get f′(a) = = .This simplifies to f′(a) = . Now we can factor out the h from the numeratorand cancel it with the h in the denominator: f′(a) = (40a + 20h). Nowwe take the limit to get f′(a) = (40a + 20h) = 40a.7. 54
We find the derivative of a function, f(x), using the definition of the derivative, which is: f′(x) =. Here f(x) = 2x^3 and x = −3. This means that f(−3) = 2(−3)^3 = −54 and
f(−3 + h) = 2(−3 + h)^3 = 2((−3)^3 + 3(−3)^2 h + 3(−3)h^2 + h^3 ) = −54 + 54h − 18h^2 + 2h^3.If we now plug these into the definition of the derivative, we get f′(−3) = = . This simplifies to f′(−3) = . Now we can factor out the h from the numerator and cancel it with