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Determine the derivative of π of π‘ equals π‘ sin of five ππ‘.
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We have a function in π‘, which is actually the product of two functions.
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We could say that one of the functions is π‘.
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And the other is sin of five ππ‘.
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So how do we find the derivative of the product of two functions?
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For two functions π’ and π£ in π₯, the derivative of π’ times π£ is π’ times the derivative of π£ with respect to π₯ plus π£ times the derivative of π’ with respect to π₯.
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Now, of course, our function is in terms of π‘.
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So we change this slightly to the derivative of π’π£ with respect to π‘.
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And this means we can let π’ be equal to π‘ because thatβs the first function in our equation.
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And we can let π£ be equal to sin of five ππ‘ because thatβs the second function of π‘.
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We can see that weβre going to need to differentiate both of these with respect to π‘.
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If we differentiate π’ with respect to π‘, we simply get one.
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But differentiating π£ with respect to π‘ is a little bit trickier.
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We could use the chain rule.
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But we donβt need to.
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We can apply a general result.
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And that is if we differentiate sin of some constant of π‘, we get that constant multiplied by cos of that constant of π‘.
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So in this case, the derivative of sin of five ππ‘ is five π multiplied by cos of five ππ‘.
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All thatβs left is to substitute this back into our equation for the product rule.
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π’ multiplied by ππ£ ππ‘ is π‘ multiplied by five π cos five ππ‘.
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And π£ multiplied by ππ’ ππ‘ is sin five ππ‘ multiplied by one, which simplifies to five ππ‘ multiplied by cos five ππ‘ plus sin of five ππ‘.