Question

How do you do this problem? The question is attached as an image.

Answer 1

for the derivative i got -2cos(x) + 5sin(x) / (2sin(x) + 5cos(x) (2sin(x) + 5cos(x))

Answer 2

and i think you have to plug in 1/5 into the x but i keep getting the wrong answer

Answer 3

I would raise the denominator to the -1 power and use the power rule combined with the chain rule. \[y=\frac{1}{2\sin(x)+5\cos(x)} = (2\sin(x)+5\cos(x))^{-1}\]

Answer 4

Here for my derivative, I got:\[y' = -(2\sin(x)+5\cos(x))^{-2}\cdot (2\cos(x)-5\sin(x))\]

Answer 5

You don't have to really format your function to look proper till you're finding the tangent line, so in this derivative I would plug in my points \((0~,~ 1/5)\)

Answer 6

\[y' = -(2\sin(0)+5\cos(0))^{-2} \cdot (2\cos(0)-5\sin(0))\]

Answer 7

That will be your slope.

Answer 8

then use \(y-y_1 = \color{red}{y'}(x-x_1)\) and your points \((0~,~ 1/5)\) to find the equation of your tangent line in slope intercept form: \(y=(y')x+b\)

Answer 9

okay thank you very much!! i got m = -0.08 and y = 1/5

Answer 10

I got m = -0.08 also, or rather \(-\dfrac{2}{25}\)

\[y-\frac{1}{5} = -\frac{2}{25}(x-0)\]

Answer 11

Now just put it in slope intercept form. \[y=mx+b\]

Answer 12

y = -0.08x + 1/5

Answer 13

\[\checkmark\]