Question

f(x)=[x+csc(x^3+3)]^-3 So far for f '(x) I have:
-3[x+csc(x^3+3)]^-4[1+csc(3x^2)+(x^3+3)(-csccot)]
The answer I'm given is:
-3[x+csc(x^3+3)]^-4[1-3x^2csc(x^3+3)cot(x^3+3)]

Answer 1

What am I doing wrong? Do you not do the chain rule on csc(x^3+3) ?

Answer 2

what is cscot?

Answer 3

@zzr0ck3r It's csc cot , isn't - csc cot the derivative of csc?

Answer 4

ok, we know that:\[\csc x = \frac{ 1 }{ \sin x }\]
now we have:
\[f(x) = \left[ x + \frac{ 1 }{ \sin(x ^{3}+3) } \right]^{-3}\]
when we derivate this, we will get have to derivate the parenthesis^-3, insides of the parenthesis, fraction, sinus and insides of the sinus

\[f \prime (x) = -3*\left[ x + \frac{ 1 }{ \sin(x ^{3} + 3) } \right]^{-4} * [1 + \frac{ -\cos(x ^{3}+3)*3x ^{2} }{ \sin ^{2}(x ^{3}+3) } ]\]

the first part was solved with \[x ^{n}\prime = n*x ^{n-1}\]
the second part, insides of the parenthesis we derivated part by part; x derivated is 1;
the fraction was derivated following the rule:

\[\frac{ x }{ y }\prime = \frac{ x \prime*y - x*y \prime }{ y ^{2} }\]

in this part, x = 1 and y =sin(x^3 +3); x derivated is zero and y derivated is 3x^2*cos(x^3+3)


\[f \prime (x) = -3*\left[ x + \frac{ 1 }{ \sin(x ^{3} + 3) } \right]^{-4} * [1 + \frac{ -\cos(x ^{3}+3)*3x ^{2} }{ \sin ^{2}(x ^{3}+3) } ]\]
\[= -3*\left[ x + \frac{ 1 }{ \sin(x ^{3} + 3) } \right]^{-4} * [1 - 3x ^{2}*\frac{ \cos(x ^{3}+3) }{ \sin(x ^{3}+3)*\sin(x ^{3}+3) }]\]

we know that (cos x)/(sin x) = cot x and 1/sin x = csc x, and we can see that this is the same as your solution;)

if you have any questions feel free to ask

Answer 5

Awesome, thanks so much!