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Evaluate dπ¦ by dπ₯ at π₯ equals one if π¦ equals six π₯ squared minus two π₯ minus three to the power of negative three multiplied by π₯ squared minus two to the power of five.
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So the first thing we need to do is find dπ¦ by dπ₯ by differentiating our function.
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And whatever we get, weβll then evaluate it at π₯ equals one.
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So we want to differentiate our function π¦.
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And we can see that π¦ is actually two functions being multiplied together.
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Itβs a product, so we use the product rule.
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Letβs recall the product rule.
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The product rule says that if π¦ equals π’π£, then dπ¦ by dπ₯ equals π’ dπ£ dπ₯ plus π£ dπ’ dπ₯.
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So, for our question, our π’ will be six π₯ squared minus two π₯ minus three to the power of negative three.
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And our π£ will be π₯ squared minus two to the power of five.
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And we can see from the product rule weβre going to need dπ£ by dπ₯ and dπ’ by dπ₯.
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So letβs go ahead and find dπ’ by dπ₯.
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Now because π’ is a function of a function, weβre going to need the chain rule.
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So letβs just remember the chain rule first of all.
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The chain rule says that if π’ is π of π and π is π of π₯, then dπ’ by dπ₯ is equal to dπ’ by dπ multiplied by dπ by dπ₯.
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So letβs apply this to our function π’.
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So we have that π is six π₯ squared minus two π₯ minus three.
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And that differentiates with respect to π₯ to get that dπ by dπ₯ is equal to 12π₯ minus two.
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And we got that by multiplying the coefficient six by the power two and then taking one away from the power to get 12π₯.
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The two π₯ differentiates to give us two.
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And then the three is just a constant.
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And because we said that π is equal to six π₯ squared minus two π₯ minus three, we have that π’ is equal to π to the power of negative three.
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And so dπ’ by dπ is negative three π to the power of negative four.
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And so, applying the formula for the chain rule, we have that dπ’ by dπ₯ is equal to negative three π to the power of negative four multiplied by 12π₯ minus two.
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And remember, π, we already defined to be six π₯ squared minus two π₯ minus three.
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So by replacing π, we get negative three multiplied by six π₯ squared minus two π₯ minus three to the power of negative four multiplied by 12π₯ minus two.
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So that gives us dπ’ by dπ₯.
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And now we also need dπ£ by dπ₯.
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And, again, π£ is a function of a function.
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So weβre going to need the chain rule again.
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So, here, we have that π is π₯ squared minus two, so that dπ by dπ₯ is two π₯.
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As we defined π to be π₯ squared minus two, we have that π£ is equal to π to the power of five so that dπ£ by dπ is equal to five π to the power of four.
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And then applying the chain rule gives us that dπ£ by dπ₯ is equal to five π to the power of four multiplied by two π₯.
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And remember, again, weβre to find π to be π₯ squared minus two.
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So we can replace π in this expression, which gives us five multiplied by π₯ squared minus two to the power of four multiplied by two π₯.
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And we can multiply together the five and the two π₯ to tidy up this a little bit.
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So that gives us 10π₯ multiplied by π₯ squared minus two to the power of four.
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So now we have dπ’ by dπ₯ and dπ£ by dπ₯.
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So we can apply the product rule.
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So Iβm just going to clear some space here.
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Okay, so applying the product rule gives us this result.
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So this is dπ¦ by dπ₯.
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But, remember, in the question, we were asked to evaluate dπ¦ by dπ₯ at π₯ equals one.
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This means we need to substitute in π₯ equals one into our first derivative.
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And once weβve substituted in one, we can tidy up our answer a little bit.
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And one to the power of negative three is just one.
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And negative one to the power of four is also one.
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So this is just one times 10 times one.
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So that gives us 10.
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And negative one to the power of five is negative one.
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And one to the power of negative four is just one.
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So this is negative one times negative three times 10, which is 30.
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So weβve got 10 plus 30, which gives us 40.
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So we differentiated our function using a combination of the chain rule and the product rule.
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And then we evaluated our first derivative at the point π₯ equals one to get our final answer of 40.