find the derivative of the following function.\ny = 5^{-x}sinx\n\\frac{d}{dx}5^{-x}sinx=\\square
find the derivative of the following function.\ny = 5^{-x}sinx\n\\frac{d}{dx}5^{-x}sinx=\\square
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = u\cdot v), then (y^\prime=u^\prime v + uv^\prime). Let (u = 5^{-x}) and (v=\sin x). First, find (u^\prime): Using the formula (\frac{d}{dx}a^{kx}=ka^{kx}\ln a) (here (a = 5), (k=- 1)), so (u^\prime=\frac{d}{dx}5^{-x}=-5^{-x}\ln5). And (v^\prime=\frac{d}{dx}\sin x=\cos x).
Step2: Substitute into the product rule
(\frac{d}{dx}(5^{-x}\sin x)=u^\prime v+uv^\prime) (=-5^{-x}\ln5\cdot\sin x + 5^{-x}\cdot\cos x) (=5^{-x}(\cos x-\ln5\sin x))
Answer:
(5^{-x}(\cos x-\ln5\sin x))