the h in the denominator: f′(−3) = (54 − 18h + 2h^2 ). Now we takethe limit to get f′(−3) = (54 − 18h + 2h^2 ) = 54.- −9x^2
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here, f(x) = −3x^3 . This means that f(x + h) = −3(x + h)^3 = −3(x^3 + 3x^2 h
+ 3xh^2 + h^3 ) = −3x^3 − 9x^2 h − 9xh^2 − 3h^3.If we now plug these into the definition of the derivative, we get f′(x) = = . This simplifies to f′(x) = . Now we can factor out the h from the numerator and cancel it with
the h in the denominator: f′(x) = = (−9x^2 − 9xh − 3h^2 ). Now wetake the limit to get f′(x) = (−9x^2 − 9xh − 3h^2 ) = −9x^2.- 5 x^4
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = x^5.
This means that f(x + h) = (x + h)^5 = (x^5 + 5x^4 h + 10x^3 h^2 + 10x^2 h^3 + 5xh^4 + h^5 ). If we now plugthese into the definition of the derivative, we get f′(x) = = .
This simplifies to f′(x) = . Now we can factor outthe h from the numerator and cancel it with the h in the denominator: f′(x) = = (5x^4 + 10x^3 h + 10x^2 h^2 5 xh^3 + h^4 ). Now wetake the limit to get f′(x) = (5x^4 +10x^3 h + 10x^2 h^2 + 5xh^3 + h^4 ) = 5x^4.