Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MATH 320 Calculus II Differentiation review exercises solutions Differentiate 1. f ( x) 13x10 17 x5 19 x 23 Note that this verifies 130 x 85 x f ( x) 13 10 x9 17 5x 4 19 1x0 0 130 x9 85x 4 19 9 2. g (t ) 6 t 4 19 dx 13x10 17 x 5 19 x C . 1 6 t First, use algebra (properties of exponents) to rewrite g (t ) 6t 1 / 2 16 t 1 / 2 . g t 6 12 t 1 / 2 16 12 t 3 / 2 3t 1 / 2 121 t 3 / 2 3 1 3 1 Optional, rewrite as g t 1 / 2 or as g t . 3/ 2 t 12t t 12t t 3. y t 2 t 5t 25 t Solution 1 (not shown): use quotient rule, noting t 2 t t 5 / 2 . Solution 2: first use algebra to rewrite y t 3 / 2 5 25t 1 , then differentiate to obtain dy 3 1 / 2 dy 3 25 2 t 0 25 t 2 ; optionally rewrite as 2 t 2 . dx dx t 4. k ( x) sec x tan x Product rule uv uv uv : k x sec x sec 2 x sec x tan x tan x sec3 x sec x tan 2 x Optional: use algebra and trigonometric identity tan 2 x 1 sec 2 x to rewrite this as k x sec x sec 2 x tan 2 x sec x 1 2 tan 2 x or k x sec x 2 sec 2 x 1 . 5. h( x) e x sin( e x ) Product rule, then chain rule – if u is differentiable function of x, then e e hx e x cos e x d dx x x 2 d dx sin u cos u dudx : sin e x e x cos e x e x sin e x e 2 x cos e x e x sin e x 6. j (t ) sin 2 t cos2 t Solution 1 (from class): first use the trigonometric identity sin 2 t cos 2 t 1 , thus j (t ) 1 and jt 0 . Solution 2 (derivative of sum, then chain rule on each term; note sin 2 t sin t ): j(t ) 2 sin t dtd sin t 2 cos t dtd cos t 2 sin t cos t 2 cos t ( sin t ) 2(sin t cos t cos t sin t ) 0 2 7. m( s) 49s 19 39 Chain rule (power rule form – if u is differentiable function of x, then d dx 38 u nu n 38 38 ms 4 399s 19 dsd 9s 19 4 39 99s 19 14049s 19 n 1 du dx ): 8. n( x) 3 3 x cot x Use algebra to rewrite as n x 3 x cot x 1/ 2 nx 1 3 3x cot x 2 / 3 3 csc 2 x , then use power rule form of chain rule: 3 csc 2 x 2/3 33x cot x t2 1 9. s ln 3 t 2 t Solution 1 (first use properties of logarithms to rewrite, then use chain rule, d ln u 1 du ): dx u dx ds 1 d 2 1 d 3 2t 3t 2 2 2 t 1 t 2 t s ln t 2 1 ln t 3 2t , so dt t 1 dt t 3 2t dt t 2 1 t 3 2t Optional: use algebra to rewrite as ds 2t t 3 2t 3t 2 2 t 2 1 2t 4 4t 2 3t 4 5t 2 2 t4 t2 2 dt t 2 1 t 3 2t t 2 1 t 3 2t t 2 1 t 3 2t du Solution 2 (see #10) – use chain rule, and quotient rule to find . dt 10. z x2 1 x3 2 x 2 u vu uv Quotient rule : v2 v dz x3 2 x 2 x x 2 1 3x 2 2 and optionally, simplify to 2 dx x3 2 x dz 2 x3 4 x 2 (3x 4 5 x 2 2) ( x 4 x 2 2) . 2 2 dx x3 2 x x3 2 x t2 1 and change variable x to t in #10: t 3 2t ds 1 du 1 t4 t2 2 t4 t2 2 , the same answer we obtained in 2 2 t 1 dt u dt t 2 1 t 2 2t t 3 2t t 3 2t solution 1. Solution 2 for #9, let s = ln u, where u 11. y cos x x sin x e dy sin x x cos x sin x 0 x cos x dx 12. g ( x) x cos( x 4 ) 2x4 Use quotient rule, then for derivative of numerator, use product then chain rule: 2 x 4 dxd x cos x 4 x cos x 4 dxd 2 x 4 g x 2 2x4 2x 4 2x 4 13. Find 2x sin x cosx 8x 2x x sin x 4 4 x3 cos x 4 x cos x 4 8 x3 2 4 4x4 4 4 dy dx 4 4 cos x 4 2 : xy4 x 4 y csc x Differentiate implicitly with respect to x (on left, use difference, then product and chain): d d csc x xy4 x 4 y dx dx dy dy Now, solve for dy : x 4 y3 dx 1y 4 x4 dx 4 x3 y csc x cot x dx 3 4xy 3 x4 dy dx csc x cot x y 4 4 x3 y , dy 4 x3 y csc x cot x y 4 (for (x, y) satisfying 4 xy3 x 4 0 and both csc x 3 4 dx 4 xy x and cot x are defined. so End notes: Similarly as in #1, we have verified the indefinite integral of our answer equals the given function plus a constant, C, of integration. Using integration rules in section 5.1, we could evaluate 130 x9 85 x 4 19 dx. , the answer in problem 1. If we first used algebra, or a trigonometric identity, we could integrate the “answers” in problems 2 and 3. The intent of problem 6 is to advise you to be on the lookout for using identities. To integrate functions where the chain rule is used to check the answer, we’ll later use the “method of substitution”; this is relevant in problems 7, 8, and parts of problems 5 and 8. To integrate functions where the product rule is used to check the answer, we’ll later use “integration by parts”; this is relevant in problems 4, and parts of problems 5 and 11. An alternative to the quotient rule for differentiation is to write uv uv 1 , and differentiating using the product, then chain rule. Consequently, the method of substation and integration by parts may be appropriate for integrating functions where the quotient rule is used to check the answer. This is relevant in problem 10. We’ll later see “partial fractions” is relevant in integrating the answers in problems 9 and 10. We’ll later see additional integration formulas and techniques of integration; it’s unlikely we would be asked to integrate such a function.. Integrating the answer from problem 12 would entail using a number of special tricks. The general message is that corresponding to differentiation rules that one uses to check an antiderivative answer are appropriate integration rules and / or integration techniques. 4