Differentiating Trig Functions (1 Viewer)

Menomaths · Jul 11, 2013

Hello, I am stuck on these questions;
Differentiate
1. (sinx - cosx)^2
2. tan (pi - x)
3. tanx(degrees)

asianese · Jul 11, 2013

1. Use the chain rule. For example, how do you differentiate (5x^4-2x)^2?

2. Again, use the chain rule. Differentiate the outside function, and then the inside function.

3. Convert x degrees to something radians. Then differentiate using the chain rule again.

bedpotato · Jul 11, 2013

1. Use the chain rule. Bring the power down, subtract 1 from the power. Differentiate what's inside the brackets, and multiply the whole equation by it.
y = (sinx)^3
y' = 3(sinx)^2(cosx)
= 3sin^2xcosx

2. If you have y = tan(2x), then dy/dx would be 2sec^2(2x). Differentiate what's inside the brackets, bring it out to the front.
so y = tan(pi - x). Differentiate what's inside the brackets, you get -1. Place -1 at the front.
y' = -sec^2(pi - x)

3. Convert degrees to radians. and follow the steps in no. 2. If it was 90 degrees, that would be pi/2.
y = tan(pi/2 x)
y' = pi/2sec^2(pi/2 x)

Menomaths · Jul 11, 2013

asianese said:
1. Use the chain rule. For example, how do you differentiate (5x^4-2x)^2?

2. Again, use the chain rule. Differentiate the outside function, and then the inside function.

3. Convert x degrees to something radians. Then differentiate using the chain rule again.

I know I have to use chain rule for all of them, but my answers are wrong.

Menomaths · Jul 11, 2013

bedpotato said:
1. Use the chain rule. Bring the power down, subtract 1 from the power. Differentiate what's inside the brackets, and multiply the whole equation by it.
y = (sinx)^3
y' = 3(sinx)^2(cosx)
= 3sin^2xcosx

2. If you have y = tan(2x), then dy/dx would be 2sec^2(2x). Differentiate what's inside the brackets, bring it out to the front.
so y = tan(pi - x). Differentiate what's inside the brackets, you get -1. Place -1 at the front.
y' = -sec^2(pi - x)

3. Convert degrees to radians. and follow the steps in no. 2. If it was 90 degrees, that would be pi/2.
y = tan(pi/2 x)
y' = pi/2sec^2(pi/2 x)

The answers are, for
1. 2(sinx-cosx)(sinx+cosx)
2.-sec^2x (possible misprint)
3.(pi/180) sec^2 x(degrees)

bedpotato · Jul 11, 2013

Menomaths said:
All of your answers are wrong, except for 2 I think because even I got what you're getting so it could be a misprint in the book
The answers are, for
1. 2(sinx-cosx)(sinx+cosx)
2.-sec^2x (possible misprint)
3.(pi/180) sec^2 x(degrees)

For 1 and 3, I wasn't giving you the answers, just examples.

asianese · Jul 11, 2013

No, the answers are correct (ithink).

1. (sinx - cosx)^2
2. tan (pi - x)
3. tanx(degrees)

$\begin{align*} \dfrac{d}{dx}(\sin x -\cos x)^2 &= 2(\sin x - \cos x) \cdot \dfrac{d}{dx} ( \sin x - \cos x) \\&= 2(\sin x - \cos x) \cdot ( \cos x + \sin x) \end{align*}$

$\begin{align*} \dfrac{d}{dx} \tan (\pi -x ) &= \sec^2 (\pi - x) \cdot -1 \\&= -(\sec(\pi-x))^2 \\&= -(-\sec(x))^2 \\& = -\sec^2x \end{align*}$
$$ since secant is a periodic function with period $ \pi. $ think of $ \cos(\pi-x) = -\cos x$

$\begin{align*} \dfrac{d}{dx} \tan (x^{\circ}) &= \dfrac{d}{dx} \tan((\pi/180)x^c) \\&=\sec^2(x^{\circ})\cdot \dfrac{\pi}{180}. \end{align*}$

$$NB.$ 180^{\circ} = \pi ^c \Rightarrow 1^{\circ} = \dfrac{\pi}{180}^c.$

bedpotato · Jul 11, 2013

Menomaths said:
I know I have to use chain rule for all of them, but my answers are wrong.

What's your working out for 1 and 3?

Menomaths · Jul 11, 2013

bedpotato said:
For 1 and 3, I wasn't giving you the answers, just examples.

Yeah just realized so I edited it haha, didn't refresh page so didn't know you posted this

asianese · Jul 11, 2013

Yeah sure. You end up with the same result.

bedpotato · Jul 11, 2013

Yeah but it takes longer to differentiate, because you have to use both the chain rule and product rule. Using the chain rule without expanding is faster.

Squar3root · Jul 11, 2013

Answers:
1) 2(sinx -cosx)(sinx + cosx) = -2cos(2x) simple chain rule, if you cant do that, let u= sinx -cosx and go from there and then expand brackets and clean up and simplify

2) tan(pi -x) = -tanx (using ASTC) therefore dy/dx = -(secx)^2

3) tan(xdegrees) = tan(pi*x/180) therfore dy/dx = (pi/180)*[sec(pi*x/180)^2]

Menomaths · Jul 11, 2013

asianese said:
No, the answers are correct (ithink).

1. (sinx - cosx)^2
2. tan (pi - x)
3. tanx(degrees)

$\begin{align*} \dfrac{d}{dx}(\sin x -\cos x)^2 &= 2(\sin x - \cos x) \cdot \dfrac{d}{dx} ( \sin x - \cos x) \\&= 2(\sin x - \cos x) \cdot ( \cos x + \sin x) \end{align*}$

$\begin{align*} \dfrac{d}{dx} \tan (\pi -x ) &= \sec^2 (\pi - x) \cdot -1 \\&= -(\sec(\pi-x))^2 \\&= -(-\sec(x))^2 \\& = -\sec^2x \end{align*}$
$$ since secant is a periodic function with period $ \pi. $ think of $ \cos(\pi-x) = -\cos x$

$\begin{align*} \dfrac{d}{dx} \tan (x^{\circ}) &= \dfrac{d}{dx} \tan((\pi/180)x^c) \\&=\sec^2(x^{\circ})\cdot \dfrac{\pi}{180}. \end{align*}$

$$NB.$ 180^{\circ} = \pi ^c \Rightarrow 1^{\circ} = \dfrac{\pi}{180}^c.$

I understand cos(pi-x) = -cosx but how does that make -(sec(pi-x))^2 = -(-sec(x))^2

asianese · Jul 11, 2013

Since $\sec x = \dfrac{1}{\cos x} $ and $\dfrac{1}{\cos ( \pi -x)} = -\dfrac{1}{\cos x}$ Assuming that neither denominator is 0. Then you consider $\sec(\pi -x) = -\sec(x)$ and substitute this into the bracket.

Menomaths · Jul 11, 2013

Ah, Thanks for the help

Menomaths · Jul 12, 2013

I have another question and I don't want to spam threads so I'll just ask my question here, The region under the curve y=tan x between x=pi/6 and x=pi/3 is rotated about the x-axis. What is the volume of the solid of revolution. Use the face that tan^2 x = sec^2 x-1. Attempted it a few times but keep on getting the same wrong answer

.

HeroicPandas · Jul 12, 2013

Menomaths said:
I have another question and I don't want to spam threads so I'll just ask my question here, The region under the curve y=tan x between x=pi/6 and x=pi/3 is rotated about the x-axis. What is the volume of the solid of revolution. Use the face that tan^2 x = sec^2 x-1. Attempted it a few times but keep on getting the same wrong answer .

$V = \pi \int_a^b y^2 dx \\ \\ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left ( tan^2x \right )dx \\ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left ( sec^2x-1 \right )dx \\ V = \pi \left [ tanx - x \right ]_{\pi/6}^{\pi/3} \\ V = \pi \left [ \left ( \sqrt{3} - \frac{\pi}{3} \right )-\left ( \frac{1}{\sqrt{3}}-\frac{\pi}{6} \right ) \right ]$

Menomaths · Jul 12, 2013

HeroicPandas said:
$V = \pi \int_a^b y^2 dx \\ \\ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left ( tan^2x \right )dx \\ V = \pi \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left ( sec^2x-1 \right )dx \\ V = \pi \left [ tanx - x \right ]_{\pi/6}^{\pi/3} \\ V = \pi \left [ \left ( \sqrt{3} - \frac{\pi}{3} \right )-\left ( \frac{1}{\sqrt{3}}-\frac{\pi}{6} \right ) \right ]$

Oh yeah, gotta integrate the -1 to, haha thanks.

HeroicPandas · Jul 12, 2013

Menomaths said:
Oh yeah, gotta integrate the -1 to, haha thanks.

no problemo

Menomaths · Jul 13, 2013

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Differentiating Trig Functions (1 Viewer)

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