find the equation of the tangent line to the curve $y = \\frac{-7}{\\sin x+\\cos x}$ at the point $(0…

Explanation:

Step1: Differentiate using quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = - 7$, $u^\prime=0$, $v=\sin x+\cos x$, and $v^\prime=\cos x-\sin x$. So, $y^\prime=\frac{0\times(\sin x+\cos x)-(-7)\times(\cos x - \sin x)}{(\sin x+\cos x)^{2}}=\frac{7(\cos x - \sin x)}{(\sin x+\cos x)^{2}}$.

Step2: Find the slope at the given point

Substitute $x = 0$ into $y^\prime$. When $x = 0$, $\sin(0)=0$ and $\cos(0)=1$. Then $y^\prime(0)=\frac{7(\cos(0)-\sin(0))}{(\sin(0)+\cos(0))^{2}}=\frac{7(1 - 0)}{(0 + 1)^{2}}=7$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,-7)$ and $m = 7$. Substituting these values, we get $y-(-7)=7(x - 0)$.

Step4: Simplify the equation

$y + 7=7x$, so $y=7x-7$.

Answer:

$y = 7x-7$