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If π¦ equals negative eight sin of sin of six π₯ minus cos of sin of six π₯, find dπ¦ by dπ₯.
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We can differentiate this term-by-term.
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However, each of our terms is a function of a function.
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And when we want to differentiate a function of a function, we use the chain rule.
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Letβs remind ourselves of the chain rule.
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This says that if π¦ equals π of π’ and π’ equals π of π₯, then dπ¦ by dπ₯ equals dπ¦ by dπ’ multiplied by dπ’ by dπ₯.
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Letβs call this first term π§.
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And to avoid confusion with the π¦ in the question and the π¦ in the chain rule formula, letβs replace π¦ in the formula with π§.
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We let our inner function π’ equal sin of six π₯.
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And so, π§ equals negative eight sin of π’.
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To use the chain rule formula, we need dπ§ by dπ’ and dπ’ by dπ₯.
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Letβs start by finding dπ§ by dπ’.
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This is going to be the derivative of negative eight sin of π’ with respect to π’.
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To do this, we remember the main derivatives of trigonometric functions.
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And we see that the derivative of sin of π₯ is cos of π₯.
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And so, the derivative of negative eight sin of π’ with respect to π’ is negative eight cos of π’.
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And now, we find dπ’ by dπ₯.
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This is the derivative of sin of six π₯ with respect to π₯.
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If we use the fact that the derivative of sin of ππ₯ equals π cos of ππ₯, then dπ’ by dπ₯ equals six cos of six π₯.
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And now, we can apply the formula for the chain rule.
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Negative eight sin of sin of six π₯ differentiates to negative eight cos of π’ multiplied by six cos of six π₯.
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We remember that we let π’ equal sin of six π₯.
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And so, we can replace π’ with sin of six π₯.
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And now, we move on to our next term.
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Letβs call this function β.
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And so, we replace π§ in our formula for the chain rule with β.
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And weβll replace π’ with π€ because we already assigned π’ in our previous term.
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So, π€ is our inner, function sin of six π₯, which means that β equals cos of π€.
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We need to find dβ by dπ€ and dπ€ by dπ₯.
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Letβs start by finding dβ by dπ€.
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This is the derivative of cos of π€ with respect to π€.
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We can use the fact that the derivative of cos of π₯ with respect to π₯ is negative sin of π₯.
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So, dβ by dπ€ equals negative sin of π€.
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We also need to find dπ€ by dπ₯.
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This is the derivative of sin of six π₯ with respect to π₯.
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Using the fact that the derivative of sin of ππ₯ with respect to π₯ is π cos of ππ₯, dπ€ by dπ₯ equals six cos of six π₯.
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And so, cos of sin of six π₯ differentiates to negative sin of π€ multiplied by six cos of six π₯.
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We remember that we let π€ equal sin of six π₯.
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And so, we replace π€ with sign of six π₯.
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And now, weβve seen what each term differentiates to, we can put it together to get dπ¦ by dπ₯. dπ¦ by dπ₯ equals negative eight cos of sin of six π₯ multiplied by six cos of six π₯ minus, because it was a minus in the question, negative sin of sin of six π₯ multiplied by six cos of six π₯.
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Here, we are subtracting a negative, so we can just write this as an add.
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And six cos of six π₯ appears in both terms, so we can take this out as a common factor.
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This leaves us with six cos of six π₯ all multiplied by negative eight cos of sin of six π₯ add sin of sin of six π₯.