find \\frac{dy}{dx}.\ny = 4\\sin x\\tan x\n\\frac{dy}{dx}=\\square
find \\frac{dy}{dx}.\ny = 4\\sin x\\tan x\n\\frac{dy}{dx}=\\square
Answer
Explanation:
Step1: Apply the product rule
The product rule states that if (y = uv), then (y^\prime=u^\prime v + uv^\prime). Let (u = 4\sin x) and (v=\tan x). First, find (u^\prime) and (v^\prime). (u^\prime=\frac{d}{dx}(4\sin x)=4\cos x) (using the rule (\frac{d}{dx}(\sin x)=\cos x) and the constant - multiple rule (\frac{d}{dx}(cf(x)) = c\frac{d}{dx}(f(x)))). (v^\prime=\frac{d}{dx}(\tan x)=\sec^{2}x) (using the rule (\frac{d}{dx}(\tan x)=\sec^{2}x)).
Step2: Substitute into the product rule formula
By the product rule (y^\prime = u^\prime v+uv^\prime). Substitute (u = 4\sin x), (u^\prime = 4\cos x), (v=\tan x), and (v^\prime=\sec^{2}x) into the formula: (y^\prime=4\cos x\tan x+4\sin x\sec^{2}x). Since (\tan x=\frac{\sin x}{\cos x}), then (4\cos x\tan x = 4\sin x). And (\sec x=\frac{1}{\cos x}), so (4\sin x\sec^{2}x=\frac{4\sin x}{\cos^{2}x}). Another way to simplify: [ \begin{align*} y^\prime&=4\cos x\frac{\sin x}{\cos x}+4\sin x\frac{1}{\cos^{2}x}\ &=4\sin x+\frac{4\sin x}{\cos^{2}x}\ &=4\sin x\left(1 + \frac{1}{\cos^{2}x}\right)\ &=4\sin x\frac{\cos^{2}x + 1}{\cos^{2}x}\ \end{align*} ] Or factor out (4\sin x): (y^\prime=4\sin x(1+\sec^{2}x))
Answer:
(4\sin x + \frac{4\sin x}{\cos^{2}x}) (or (4\sin x(1+\sec^{2}x)) or (4\sin x+\ 4\sin x\sec^{2}x))