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Zorluk: OrtaDifferentiation of Trigonometric, Exponential, and Logarithmic Functions

If y=ln⁡(cos⁡(3x))y = \ln(\cos(3x)), what is dydx\frac{dy}{dx}?

  1. −3tan⁡(3x)-3\tan(3x)Cevap
  2. B
    3tan⁡(3x)3\tan(3x)
  3. C
    −tan⁡(3x)-\tan(3x)
  4. D
    −3cot⁡(3x)-3\cot(3x)

Cevap

−3tan⁡(3x)-3\tan(3x)
Differentiating y=ln⁡(cos⁡(3x))y = \ln(\cos(3x)) requires using the chain rule twice: first for ln⁡(u)\ln(u), giving 1cos⁡(3x)\frac{1}{\cos(3x)}, and then for cos⁡(3x)\cos(3x), giving −3sin⁡(3x)-3\sin(3x). Multiplying these results yields −3sin⁡(3x)cos⁡(3x)\frac{-3\sin(3x)}{\cos(3x)}, which simplifies to −3tan⁡(3x)-3\tan(3x).

Adım Adım Çözüm

1
Apply the chain rule formula for differentiating logarithmic functions ddx[ln⁡(u)]=1ududx\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}
dydx=1cos⁡(3x)⋅ddx(cos⁡(3x))\frac{dy}{dx} = \frac{1}{\cos(3x)} \cdot \frac{d}{dx}(\cos(3x))
The outer function is ln⁡(u)\ln(u) where u=cos⁡(3x)u = \cos(3x).
2
Differentiate the inner trigonometric function cos⁡(3x)\cos(3x)
ddx(cos⁡(3x))=−3sin⁡(3x)\frac{d}{dx}(\cos(3x)) = -3\sin(3x)
By the chain rule, ddx[cos⁡(kx)]=−ksin⁡(kx)\frac{d}{dx}[\cos(kx)] = -k\sin(kx).
3
Multiply the terms and simplify using trigonometric identities
dydx=−3sin⁡(3x)cos⁡(3x)=−3tan⁡(3x)\frac{dy}{dx} = \frac{-3\sin(3x)}{\cos(3x)} = -3\tan(3x)
Using the identity sin⁡(θ)cos⁡(θ)=tan⁡(θ)\frac{\sin(\theta)}{\cos(\theta)} = \tan(\theta), the derivative simplifies to −3tan⁡(3x)-3\tan(3x).

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule
Tahmini Süre:1m 30s
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