# Problem with trigonomic derivative (pretty simple)

• Mar 6th 2009, 06:19 PM
vane505
Problem with trigonomic derivative (pretty simple)
I know how to derivate sinx, cosx and standard equations like

y = x^4 + sqrt.(x)
y' = 4x^3 + x^-1/2

y = sinx + cosx
y' = cosx - sinx

and so on. But now they want me to "decide an equation for the tangent to the curve":

a) y = cosx (pi/2, 0)

Answer: y = -x + pi/2

b) y = 2x - 5sinx (pi, 2pi)

I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

Appreciate help!!
• Mar 6th 2009, 06:30 PM
Prove It
Quote:

Originally Posted by vane505
I know how to derivate sinx, cosx and standard equations like

y = x^4 + sqrt.(x)
y' = 4x^3 + x^-1/2

y = sinx + cosx
y' = cosx - sinx

and so on. But now they want me to "decide an equation for the tangent to the curve":

a) y = cosx (pi/2, 0)

Answer: y = -x + pi/2

b) y = 2x - 5sinx (pi, 2pi)

I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

Appreciate help!!

I take it that the numbers in the brackets are the co-ordinates of the point the tangent should touch...

a) $\displaystyle \frac{dy}{dx} = -\sin{x}$.

At the point $\displaystyle x = \frac{\pi}{2}, \frac{dy}{dx} = -\sin{\frac{\pi}{2}} = -1$.

Remember that the equation for a line is given by $\displaystyle y = mx + c$ where $\displaystyle m$ is the slope and $\displaystyle c$ is the $\displaystyle y$-intercept. Since we have the slope of the tangent line and a point that lies on the line, we can solve for $\displaystyle c$.

$\displaystyle y = mx + c$

$\displaystyle 0 = -1\left(\frac{\pi}{2}\right) + c$

$\displaystyle 0 = -\frac{\pi}{2} + c$

$\displaystyle c = \frac{\pi}{2}$.

So the equation of the tangent line to the curve at the point given is

$\displaystyle y = -x + \frac{\pi}{2}$.
• Mar 6th 2009, 06:33 PM
Prove It
Quote:

Originally Posted by vane505
I know how to derivate sinx, cosx and standard equations like

y = x^4 + sqrt.(x)
y' = 4x^3 + x^-1/2

y = sinx + cosx
y' = cosx - sinx

and so on. But now they want me to "decide an equation for the tangent to the curve":

a) y = cosx (pi/2, 0)

Answer: y = -x + pi/2

b) y = 2x - 5sinx (pi, 2pi)

I'm new to derivatives with sinx, and they have almost no examples in the book unfortunetly. I have almost no idea how to solve this, except that I guess pi would stand for 180 degrees, so then pi/2 would be x = 90 degrees. After that I'm lost!

Appreciate help!!

b) $\displaystyle y = 2x - 5\sin{x}$ so $\displaystyle \frac{dy}{dx} = 2 - 5\cos{x}$.

At the point $\displaystyle x = \pi, \frac{dy}{dx} = 2 - 5\cos{\pi} = 2 - 5(-1) = 7$. This is $\displaystyle m$.

So now we put $\displaystyle x, y, m$ into $\displaystyle y = mx + c$ and solve for $\displaystyle c$.

$\displaystyle y = mx + c$

$\displaystyle 2\pi = 7\pi + c$

$\displaystyle c = -5\pi$.

So the equation of the tangent line at the point given is

$\displaystyle y = 7x - 5\pi$.
• Mar 6th 2009, 07:38 PM
vane505
Great! Thank you very much for your help and very clear explenation, I get it now :)