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If π¦ is equal to one plus two times the sin squared of two π₯ all divided by three minus three sin squared of two π₯, find dπ¦ by dπ₯.
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Weβre given π¦ as the quotient of two trigonometric functions and asked to find dπ¦ by dπ₯.
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Thatβs the derivative of π¦ with respect to π₯.
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So weβll do this by using the quotient rule.
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And we could do this directly right now.
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However, itβs always worth checking if we can simplify our expression.
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And in this case, we can.
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Weβll take out the common factor of three in our denominator.
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This would give us a new denominator of three times one minus the sin squared of two π₯.
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And we can simplify one minus the sin squared of two π₯ by using the Pythagorean identity.
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Recall, the Pythagorean identity tells us the cos squared of π plus the sin squared of π is equivalent to one.
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Weβll subtract the sin squared of π from both sides of this equivalence.
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This gives us the cos squared of π is equivalent to one minus the sin squared of π.
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And we can make this exactly what we have in our denominator.
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We just need to set π equal to two π₯.
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So instead of having one minus the sin squared of two π₯ in our denominator, we can instead have the cos squared of two π₯.
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So by using the Pythagorean identity, we can rewrite π¦ as one plus two sin squared of two π₯ all divided by three times the cos squared of two π₯.
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And now we could differentiate this by using the power rule for differentiation.
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However, we can actually make this even easier.
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Weβre going to divide both terms in our numerator separately by our denominator.
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Doing this gives us the following expression.
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And we can simplify both of these by using trigonometric identities.
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First, weβre dividing by the cos squared in our first term.
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And we need to remember dividing by the cos of π is the same as the sec of π.
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So we can rewrite this as one-third times the sec squared of two π₯.
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We can do something very similar in our second term.
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Remember, sine divided by cosine is the tangent.
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So the sin squared of two π₯ divided by the cos squared of two π₯ will be the tan squared of two π₯.
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So we can rewrite the second term as two-thirds times the tan squared of two π₯.
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And now we could differentiate this term by term by using either the product rule, the chain rule, or the general power rule.
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And any of these methods would work and give us the correct answer.
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However, we can actually simplify this even further.
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We need to go back to our Pythagorean identity.
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We want to use the Pythagorean identity to rewrite the tan squared of two π₯.
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So weβre going to divide our Pythagorean identity through by the cos squared of π.
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Dividing the Pythagorean identity through by the cos squared of π, we get the cos squared of π divided by the cos squared of π plus the sin squared of π over the cos squared of π is equivalent to one over the cos squared of π.
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And we can simplify each of these terms.
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First, the cos squared of π over the cos squared of π is equal to one.
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Next, remember, the sin of π divided by the cos of π is the tan of π.
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So the sin squared of π divided by the cos squared of π is the tan squared of π.
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And finally, we need to remember the sec squared of π is equivalent to dividing through by the cos squared of π.
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So this gives us one plus the tan squared of π is equivalent to the sec squared of π.
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We want to rearrange this to find an expression for the tan squared of two π₯.
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So weβll start by subtracting one from both sides, giving us the tan squared of π is equivalent to the sec squared of π minus one.
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Then we just need to replace π with two π₯.
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So now we can replace the tan squared of two π₯ with the sec squared of two π₯ minus one.
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So by replacing the tan squared of two π₯ in our expression for π¦, we get that π¦ is equal to one-third times the sec squared of two π₯ plus two-thirds times the sec squared of two π₯ minus one.
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And now we can see why this is useful.
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We just distribute two-thirds over our parentheses.
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This gives us the following expression.
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And then we can combine one-third times the sec squared of two π₯ plus two-thirds times the sec squared of two π₯ to just be the sec squared of two π₯.
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So after all of this simplification, we were able to rewrite π¦ as the sec squared of two π₯ minus two-thirds.
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And this is a much easier expression to differentiate than our original expression.
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Now all thatβs left to do is differentiate this with respect to π₯.
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Of course, the derivative of the constant negative two-thirds will be equal to zero.
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So we only need to differentiate the sec squared of two π₯ with respect to π₯.
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And we have a few choices of how we want to do this.
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We could use the product rule, the chain rule, or the general power rule.
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Any method would work.
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Weβll do this by using the general power rule.
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Recall, this tells us for a differentiable function π of π₯ and constant π, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one.
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We want to use this to differentiate the sec squared of two π₯.
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So our exponent π will be equal to two, and our function π of π₯ will be the sec of two π₯.
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However, to use the general power rule, we still need to find an expression for π prime of π₯.
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Thatβs the derivative of the sec of two π₯ with respect to π₯.
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And to do this, we need to recall one of our standard trigonometric derivative results.
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For any real constant π, the derivative of the sec of ππ₯ with respect to π₯ is equal to π times the tan of ππ₯ multiplied by the sec of ππ₯.
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In our case, the value of π is equal to two.
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So we get π prime of π₯ is equal to two tan of two π₯ multiplied by the sec of two π₯.
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Weβre now ready to find an expression for dπ¦ by dπ₯.
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Remember, we showed π¦ is equal to the sec squared of two π₯ minus two-thirds.
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So dπ¦ by dπ₯ will be the derivative of this with respect to π₯.
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And as we said, weβre going to differentiate the sec squared of two π₯ by using the general power rule.
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Itβs equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one.
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And of course, we know the derivative of the constant negative two-thirds is equal to zero.
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So we can just substitute in our value of π is two and our expressions for π of π₯ and π prime of π₯.
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This gives us dπ¦ by dπ₯ is equal to two times two tan of two π₯ multiplied by the sec of two π₯ times the sec of two π₯ raised to the power of two minus one.
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And we can simplify this.
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First, our exponent of two minus one is equal to one.
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But then we can see we have the sec of two π₯ multiplied by the sec of two π₯, which we can simplify to give us the sec squared of two π₯.
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Finally, our coefficient two times two is equal to four.
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And this gives us our final answer.
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Therefore, by simplifying our expression for π¦ by using the Pythagorean identity and then applying the general power rule, we were able to show if π¦ is equal to one plus two sin squared of two π₯ all divided by three minus three sin squared of two π₯, then dπ¦ by dπ₯ would be equal to four tan of two π₯ times the sec squared of two π₯.