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# differentiation

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If f(x)=(x^3+2 cos(x))/(2sin(x)), ﬁnd f'(π/2)

Guest Oct 25, 2015

#3
+26484
+5

Both Melody and Omi have produced the correct result for f'(pi/2)  (Omi's 6pi^2/16 can be simplifird to 3pi^2/8)

(but Omi, you should have written f'(pi/2) for your last three f's, not f'(x))

.

Alan  Oct 25, 2015
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#1
+91775
+5

$$f(x)=\frac{x^3+2 cos(x)}{2sin(x)}\\ f'(x)=\frac{(2sinx)(3x^2-2sin(x))-2cos(x)(x^3-2cos(x))}{4sin^2(x)}\\ f'(x)=\frac{(2sinx)(3x^2-2sin(x))}{4sin^2(x)}-\frac{2cos(x)(x^3-2cos(x))}{4sin^2(x)}\\ f'(x)=\frac{(3x^2-2sin(x))}{2sin(x)}-\frac{cos(x)(x^3-2cos(x))}{2sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)-2cos^2(x)}{2sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-\frac{cos^2(x)}{sin^2(x)}\\ f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-(tan(x))^{-2}\\$$

$$f '(\pi/2) = \frac{3*\frac{\pi^2}{4}}{2}-1-\frac{\frac{\pi^3}{2^3}*0}{2}-0\\ f '(\pi/2) = \frac{3\pi^2}{8}-1\\ f '(\frac{\pi}{2}) = \frac{3\pi^2-8}{8}$$

$$\qquad\frac{d}{dx}\;\frac{x^3cos(x)}{2sin^2(x)}\\ \qquad=\frac{(2sin^2(x))*[3x^2cos(x)-x^3sin(x)]-4sin(x)cos(x)*x^3cos(x)}{4sin^2(x)}\\ \qquad=\frac{[6x^2sin^2(x)cos(x)-2x^3sin^3(x)]-4x^3sin(x)cos^2(x)}{4sin^2(x)}\\ \qquad=\frac{3x^2sin(x)cos(x)-x^3sin^2(x)-2x^3cos^2(x)}{2sin(x)}\\ \qquad=\frac{3x^2sin(x)cos(x)}{2sin(x)}-\frac{x^3sin^2(x)}{2sin(x)}-\frac{2x^3cos^2(x)}{2sin(x)}\\ \qquad=\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\\$$

$$f'(x)=\frac{3x^2}{2sin(x)}-1-\frac{x^3cos(x)}{2sin^2(x)}-(tan(x))^{-2}\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\left[\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\right]+2(tan(x))^{-3}(sec(x))^2\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\left[\frac{3x^2cos(x)}{2}-\frac{x^3sin(x)}{2}-\frac{x^3cos^2(x)}{sin(x)}\right]+\frac{2cos^3(x)}{sin^3(x)cos^2(x)}\\ f''(x)=\frac{12xsin(x)-6x^2cos(x)}{4sin^2(x)}-\frac{3x^2cos(x)}{2}+\frac{x^3sin(x)}{2}+\frac{x^3cos^2(x)}{sin(x)}+\frac{2cos(x)}{sin^3(x)}\\ f''(\frac{\pi}{2})=\frac{6\pi *1-0}{4*1}-\frac{3x^2*0}{2}+\frac{\pi^3}{16}+\frac{x^3*0}{1}+\frac{0}{1}\\ f''(\frac{\pi}{2})=\frac{3\pi}{2}+\frac{\pi^3}{16}\\$$

Oh dear, I thought I was supposed to find f ''  Maybe I was only supposed to find f '

oh well I will go back and do that too.

Melody  Oct 25, 2015
edited by Melody  Oct 25, 2015
#2
+8983
+5

I have differently simplified a little bit.

Omi67  Oct 25, 2015
#3
+26484
+5

Both Melody and Omi have produced the correct result for f'(pi/2)  (Omi's 6pi^2/16 can be simplifird to 3pi^2/8)

(but Omi, you should have written f'(pi/2) for your last three f's, not f'(x))

.

Alan  Oct 25, 2015
#4
+8983
0

Jetzt stimmt es aber. Danke Alan. Manchmal passe ich nicht richtig auf.

Now it is the right way.Thanks Alan.Sometimes I did not fit enough to.

Omi67  Oct 25, 2015

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