. Now we can factor out the h from the numerator and cancel it with the h in
the denominator: f′(x) = = (2x + h + 1). Now wetake the limit to get f′(x) = (2x + h + 1) = 2x + 1.- 3 x^2 + 3
We find the derivative of a function, f′(x), using the definition of the derivative, which is f′(x) =. Here f(x) = x^3 + 3x + 2 and f(x + h) = (x + h)^3 + 3(x + h) + 2 = x^3 +
3 x^2 h + 3xh^2 + h^3 + 3x + 3h + 2.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 thenumerator and cancel it with the h in the denominator: f′(x) = = (3x^2 + 3xh + h^2 + 3). Now we take the limit to get f′(x) = (3x^2 + 3xh + h^2 + 3) = 3x^2 + 3.15.
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = and f(x + h) = . If we now plug these into the
definition of the derivative, we get f′(x) = = . Notice that ifwe now take the limit, we get the indeterminate form . We cannot eliminate this problemmerely by simplifying the expression the way that we did with a polynomial. Here, wecombine the two terms in the numerator of the expression to get f′(x) = =