can you show how you arrived at your answer as well
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y = (cot(x))^7
Bring down the 7 and subtract 1 from the exponent to get:
y' = 7(cot(x))^6
But don't forget the chain rule. Derive the inside of ( ) and multiply the whole thing by it. The derivative of cot(x) is -csc²(x)
y' = -7csc²(x)(cot(x))^6
y' = -7csc²(x)cot^6(x)
The person above me has the right idea, but the derivative of cot(x) is not -csc(x)cot(x), it's -csc²(x)
Bring down the 7 and subtract 1 from the exponent to get:
y' = 7(cot(x))^6
But don't forget the chain rule. Derive the inside of ( ) and multiply the whole thing by it. The derivative of cot(x) is -csc²(x)
y' = -7csc²(x)(cot(x))^6
y' = -7csc²(x)cot^6(x)
The person above me has the right idea, but the derivative of cot(x) is not -csc(x)cot(x), it's -csc²(x)
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y = [7 cot ^6 (x)] x [ - cosec x cot x]
basically, multiply by the power... take one from the power... and times by the diff of a single cot x
basically, multiply by the power... take one from the power... and times by the diff of a single cot x
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y = cot^7(x)
Using chain rule:
dy/dx = 7*cot^6(x)*-csc²(x)
Using chain rule:
dy/dx = 7*cot^6(x)*-csc²(x)