exercises 3.5 derivatives of trigonometric functions\nscore: 27/37 answered: 14/18\nquestion 15\ntextbook…

exercises 3.5 derivatives of trigonometric functions\nscore: 27/37 answered: 14/18\nquestion 15\ntextbook videos +\nif ( f(x)=\frac{2 x^{2} \tan x}{sec x} ), find\n( f^{prime}(x)= )\nfind ( f^{prime}(2)= )\nquestion help: video message instructor\nsubmit question jump to answer

exercises 3.5 derivatives of trigonometric functions\nscore: 27/37 answered: 14/18\nquestion 15\ntextbook videos +\nif ( f(x)=\frac{2 x^{2} \tan x}{sec x} ), find\n( f^{prime}(x)= )\nfind ( f^{prime}(2)= )\nquestion help: video message instructor\nsubmit question jump to answer

Answer

Explanation:

Step1: Simplify the function

Use the trigonometric identity (\tan x=\frac{\sin x}{\cos x}) and (\sec x = \frac{1}{\cos x}). Then (f(x)=\frac{2x^{2}\tan x}{\sec x}=2x^{2}\sin x).

Step2: Apply the product rule

The product rule is ((uv)^\prime = u^\prime v+uv^\prime), where (u = 2x^{2}) and (v=\sin x).

  • Find (u^\prime): (u^\prime=(2x^{2})^\prime = 4x).
  • Find (v^\prime): (v^\prime = (\sin x)^\prime=\cos x).
  • Then (f^\prime(x)=(2x^{2})^\prime\sin x+2x^{2}(\sin x)^\prime=4x\sin x + 2x^{2}\cos x).

Step3: Evaluate (f^\prime(2))

Substitute (x = 2) into (f^\prime(x)): (f^\prime(2)=4\times2\sin2+2\times2^{2}\cos2=8\sin2 + 8\cos2).

Answer:

(f^\prime(x)=4x\sin x + 2x^{2}\cos x); (f^\prime(2)=8\sin2 + 8\cos2)