Find the 2nd Derivative f(x)=csc(x)

Problem

ƒ(x)=csc(x)

Solution

1. Identify the function ƒ(x)=csc(x) and recall the rule for the first derivative of the cosecant function.

2. Differentiate the function once to find the first derivative ƒ(x)′

d(csc(x))/d(x)=−csc(x)*cot(x)

3. Apply the product rule to find the second derivative ƒ(x)″ where the product rule is (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x) Let u=−csc(x) and v=cot(x)

ƒ(x)″=d()/d(x)*(−csc(x)*cot(x))

4. Calculate the derivatives of the individual components: (d(−)*csc(x))/d(x)=csc(x)*cot(x) and d(cot(x))/d(x)=−csc2(x)

ƒ(x)″=−csc(x)*(−csc2(x))+cot(x)*(csc(x)*cot(x))

5. Simplify the expression by multiplying the terms and combining like trigonometric functions.

ƒ(x)″=csc3(x)+csc(x)*cot2(x)

6. Factor out the common term csc(x) to reach the final simplified form.

ƒ(x)″=csc(x)*(csc2(x)+cot2(x))

Final Answer

d2(csc(x))/(d(x)2)=csc(x)*(csc2(x)+cot2(x))

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