3.6 chain rule\n5. calculate the following derivatives.\na. $\frac{d}{dx}sin(3x^{2}+5)$\nb…

Explanation:

Step1: Recall chain - rule

The chain - rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)).

Step2: Solve part A

Let (u = 3x^{2}+5), then (y=\sin(u)). (y^\prime=\cos(u)\cdot u^\prime). Since (u^\prime = 6x), (\frac{d}{dx}\sin(3x^{2}+5)=6x\cos(3x^{2}+5)).

Step3: Solve part B

Let (u=x^{5}+2x + 1), then (y = u^{4}). (y^\prime=4u^{3}\cdot u^\prime). Since (u^\prime=5x^{4}+2), (\frac{d}{dx}(x^{5}+2x + 1)^{4}=4(x^{5}+2x + 1)^{3}(5x^{4}+2)).

Step4: Solve part C

Let (u = 3x^{2}), then (y = e^{u}). (y^\prime=e^{u}\cdot u^\prime). Since (u^\prime = 6x), (\frac{d}{dx}e^{3x^{2}}=6xe^{3x^{2}}).

Step5: Solve part D

Let (u=\sin(x)), then (y = u^{5}). (y^\prime=5u^{4}\cdot u^\prime). Since (u^\prime=\cos(x)), (\frac{d}{dx}\sin^{5}(x)=5\sin^{4}(x)\cos(x)).

Step6: Solve part E

Rewrite (\frac{1}{\sqrt{x^{2}+1}}=(x^{2}+1)^{-\frac{1}{2}}). Let (u=x^{2}+1), then (y = u^{-\frac{1}{2}}). (y^\prime=-\frac{1}{2}u^{-\frac{3}{2}}\cdot u^\prime). Since (u^\prime = 2x), (\frac{d}{dx}\left(\frac{1}{\sqrt{x^{2}+1}}\right)=-\frac{x}{(x^{2}+1)^{\frac{3}{2}}}).

Step7: Solve part F

Use the product - rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x^{2}), (v=\cos^{5}(x)). Let (t=\cos(x)), then (v = t^{5}), (v^\prime=5t^{4}\cdot(-\sin(x))=- 5\cos^{4}(x)\sin(x)), (u^\prime = 2x). So (\frac{d}{dx}(x^{2}\cos^{5}(x))=2x\cos^{5}(x)-5x^{2}\cos^{4}(x)\sin(x)).

Step8: Solve part G

Let (u = \tan(2\pi x)), then (y = u^{5}). (y^\prime=5u^{4}\cdot u^\prime). Since (u^\prime=\sec^{2}(2\pi x)\cdot2\pi), (\frac{d}{dx}(\tan^{5}(2\pi x))=10\pi\tan^{4}(2\pi x)\sec^{2}(2\pi x)).

Step9: Solve part H

Rewrite (\sqrt[3]{1 + \tan(\pi x^{2})}=(1+\tan(\pi x^{2}))^{\frac{1}{3}}). Let (u = 1+\tan(\pi x^{2})), then (y = u^{\frac{1}{3}}). (y^\prime=\frac{1}{3}u^{-\frac{2}{3}}\cdot u^\prime). Let (s=\pi x^{2}), then (\tan(s)) has derivative (\sec^{2}(s)\cdot2\pi x), so (u^\prime=\sec^{2}(\pi x^{2})\cdot2\pi x). Thus (\frac{d}{dx}\sqrt[3]{1+\tan(\pi x^{2})}=\frac{2\pi x\sec^{2}(\pi x^{2})}{3(1 + \tan(\pi x^{2}))^{\frac{2}{3}}}).

Step10: Solve part I

Use the sum - rule ((u - v)^\prime=u^\prime - v^\prime) where (u = 4\cos^{7}(x)) and (v = 10\sec^{2}(x)). Let (t=\cos(x)), then (u = 4t^{7}), (u^\prime=4\times7t^{6}\times(-\sin(x))=-28\cos^{6}(x)\sin(x)), and (v = 10\sec^{2}(x)), (v^\prime=20\sec(x)\cdot\sec(x)\tan(x)=20\sec^{2}(x)\tan(x)). So (\frac{d}{dx}(4\cos^{7}(x)-10\sec^{2}(x))=-28\cos^{6}(x)\sin(x)-20\sec^{2}(x)\tan(x)).

Step11: Solve part J

Let (u=\cos(7\theta)), (v = \sin(u)), (y=\sin(v)). (y^\prime=\cos(v)\cdot v^\prime), (v^\prime=\cos(u)\cdot u^\prime), (u^\prime=-7\sin(7\theta)). So (\frac{d}{d\theta}\sin(\sin(\cos(7\theta)))=-7\sin(7\theta)\cos(\cos(7\theta))\cos(\sin(\cos(7\theta)))).

Answer:

A. (6x\cos(3x^{2}+5)) B. (4(x^{5}+2x + 1)^{3}(5x^{4}+2)) C. (6xe^{3x^{2}}) D. (5\sin^{4}(x)\cos(x)) E. (-\frac{x}{(x^{2}+1)^{\frac{3}{2}}}) F. (2x\cos^{5}(x)-5x^{2}\cos^{4}(x)\sin(x)) G. (10\pi\tan^{4}(2\pi x)\sec^{2}(2\pi x)) H. (\frac{2\pi x\sec^{2}(\pi x^{2})}{3(1 + \tan(\pi x^{2}))^{\frac{2}{3}}}) I. (-28\cos^{6}(x)\sin(x)-20\sec^{2}(x)\tan(x)) J. (-7\sin(7\theta)\cos(\cos(7\theta))\cos(\sin(\cos(7\theta))))