Question

exercises 3.5 derivatives of trigonometric functions
score: 27/37 answered: 14/18
question 15
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if ( f(x)=\frac{2 x^{2} \tan x}{sec x} ), find
( f^{prime}(x)= )
find ( f^{prime}(2)= )
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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:

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