- 4 x^3
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = x^4 . This means that f(x + h) = (x + h)^4 = (x^4 + 4x^3 h + 6x^2 h^2
+ 4xh^3 + h^4 ). 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) = = (4x^3 + 6x^2 h + 4xh^2 +h^3 ). Now we take the limit to get f′(x) = (4x^3 + 6x^2 h + 4xh^2 + h^3 ) = 4x^3.- −sin x
We find the derivative of a function, f(x), using the definition of the derivative, which is f′(x) = . Here f(x) = cos x and f(x + h) = cos(x + h). If we now plug these into
the definition of the derivative, we get f′(x) = = .Notice that if we now take the limit, we get the indeterminate form . We cannot eliminate thisproblem merely by simplifying the expression the way that we did with a polynomial. Recallthat the trigonometric formula cos(A + B) = cos A cos B − sin A sin B. Here, we can rewrite thetop expression as f′(x) = = . Wecan break up the limit into − . Next, factor cos x out of thetop of the left-hand expression: − . Now, we can break this intoseparate limits: − . The left-hand limit is = = cos x • 0 = 0. The right-hand limit is sin x = sin x • 1 = sin x. Therefore, the limit is −sin x.