Suggested Practice Problems Math 229, Calculus I Differentiate the following: Recall, for the function h(x) = f (g(x)), the chain rule is h0 (x) = f 0 (g(x)) · g 0 (x) Recall, for the function h(x) = f (x) · g(x), the product rule is h0 (x) = f (x) · g 0 (x) + f 0 (x) · g(x) f (x) , the quotient rule is g(x) g(x) · f 0 (x) − f (x) · g 0 (x) h0 (x) = (g(x))2 Recall, for the function h(x) = 1. d cos(x) sin4 (x) dx (a) Rewrite the function as cos(x) · (sin(x))4 (b) Now we differentiate using the product rule. d d d [cos(x)] · sin4 (x) cos(x) sin4 (x) = cos(x) · (sin(x))4 + dx dx dx = cos(x) · 4 (sin(x))3 · cos(x) + [− sin(x)] · sin4 (x) = 4 cos2 (x) sin3 (x) − sin5 (x) = sin3 (x) 4 cos2 (x) − sin2 (x) 2. d [tan(x2 + sin(x))] dx Use the chain rule! Recall d [tan(u)] = sec2 (u) · u0 . Note, u = x2 + sin(x). dx d d 2 tan(x2 + sin(x)) = sec2 (x2 + sin(x)) · x + sin(x) dx dx = sec2 (x2 + sin(x)) · [2x + cos(x)] d 1 3. sec dx x2 (a) First, rewrite 1 as x−2 . x2 (b) Now, we find d [sec(x−2 )] dx Note, d sec(u) = sec(u) tan(u) · u0 , where u = x−2 . dx d −2 d sec(x−2 ) = sec(x−2 ) tan(x−2 ) · x dx dx = sec(x−2 ) tan(x−2 ) · −2x−3 = −2x−3 · sec(x−2 ) · tan(x−2 ) or −2 sec(x−2 ) · tan(x−2 ) x3