Pretty much in the title but.. well im trying to find a formula for the nth time derivative of sin(x) (where the x is a position just so you know its not me wanting to find the nth derivative of sin(t) hehe).

im wondering if there is another route to finding a formula other than taking like hundreds of product rule derivatives (or using thhe leibniz formula for product rules which isn't getting me very far since its like an infinite substitution thing (if u want me to explain what i mean ask xD)) and looking for tough to spot pattern, or if its even really possible to find a formula

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TopNewestLet y = sin(x) Differentiating wrt x, y'= cos(x) = sin(π/2 +x) Similarly yn or nth derivative of sinx wrt x is sin(nπ/2 + x)

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I don't think I understand your question but do you mean this?

\[\begin{array}{l} f(x) = {\sin ^n}x\\ f'(x) = n{\sin ^{n - 1}}x\cos x \end{array}\]

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no, i mean if you take the derivative with respect to t of sin(x) n times, what would u get so for example, if you did it once, you'd get cos(x) dx/dt.. if you did another derivative, you'd get cos(x)d2x/dt2-sin(x) dx/dt... etc, im wondering if theres some way to figure out a formula for that

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nth derivative of logx(sinx) by leibiniz theorem? plz guide me

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nth derivatine of logx(sinx) by leibiniz theorem? plzzzzzzzzzz plzzzzzzzzzzzzzzz guide us

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yeah! i 'm looking for leibniz theorem

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nth derivatine of logx(sinx) by leibiniz theorem

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someone here

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