Derivative of trig Function Help

[mop969]

I have been trying to take the derivative of this trig function but I keep getting it wrong please show the steps to solve this derivative:

(csc(x))(x+cot(x))

 

[stapel]

I have been trying to take the derivative of this trig function but I keep getting it wrong

Please show your steps, so we can "see" where you're having trouble. Thank you! :wink:

 

[mop969]

Well I took the product rule of the function and got a final answer of cscx-csc^3x+(x+cotx)(cscxtanx). I dont know what is wrong?

 

[stapel]

I...got a final answer of cscx-csc^3x+(x+cotx)(cscxtanx). I dont know what is wrong?

Unfortunately, since you still haven't shown your work, there is no way for us to locate your error(s). Sorry!

At a guess, you may need to review your trig derivatives...? :wink:

 

[soroban]

$\displaystyle \text{It's hard to see your work from here . . . }$

$\displaystyle \text{But it looks like you played the }K\heartsuit\text{ instead of the }7\spadesuit$

. . $\displaystyle \text{and you transposed to the key of }F\sharp\text{ major instead of }B\flat\text{ minor.}$

 

[mop969]

Sorry my second step was cscx(1-csc^2x)+(x+cotx)(cscxtanx) This is what I got when I took the derivatives of the original function using the product rule. Please help!

 

[stapel]

Follow the link (provided earlier) to refresh your memory regarding the trig derivatives. The errors appear to have occurred there. :wink:

 

[Deleted member 4993]

I have been trying to take the derivative of this trig function but I keep getting it wrong please show the steps to solve this derivative:

(csc(x))(x+cot(x))

product rule is:

If f = u * v

then f' = u' * v + v' * u

here

u = csc(x)..........u' = - cot(x) * csc(x)

v = x + cot(x)......v' = 1 + csc[sup:3dihd6uy]2[/sup:3dihd6uy](x)

Now put those together......

 

[mop969]

I messed up with the derivatives of the trig functions. Thanks for the help everyone!

 

[BigGlenntheHeavy]

$\displaystyle f(x) \ = \ [csc(x)][x+cot(x)] \ = \ \bigg[\frac{1}{sin(x)}\bigg]\bigg[x+\frac{cos(x)}{sin(x)}\bigg]$

$\displaystyle = \ \bigg[\frac{x}{sin(x)}+\frac{cos(x)}{sin^{2}(x)}\bigg] \ = \ \frac{xsin(x)+cos(x)}{sin^{2}(x)}.$

$\displaystyle \ D_x \bigg[\frac{xsin(x)+cos(x)}{sin^{2}(x)}\bigg]=\frac{sin^{2}(x)[sin(x)+xcos(x)-sin(x)]-[xsin(x)+cos(x)][2sin(x)cos(x)]}{sin^{4}(x)}$

$\displaystyle = \ \frac{xsin^{2}(x)cos(x)-2xsin^{2}(x)cos(x)-2sin(x)cos^{2}(x)}{sin^{4}(x)} \ = \ \frac{-3xsin^{2}(x)cos(x)-2sin(x)cos^{2}(x)}{sin^{4}(x)}$

$\displaystyle = \ \frac{-3xsin(x)cos(x)-2cos^{2}(x)}{sin^{3}(x)}, We \ could \ stop \ here, \ but \ I'll \ go \ on.$

$\displaystyle = \ \frac{-3xsin(x)cos(x)}{sin^{3}(x)} \ \frac{-2cos^{2}(x)}{sin^{3}(x)} \ = \ -3xcot(x)csc(x)-2cot^{2}(x)csc(x)$

$\displaystyle = \ [ cot(x)csc(x)][-3x-2cot(x)], QED$