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Find dπ¦ by dπ₯ if π¦ is equal to the log to the base four of three π₯ squared plus two π₯ to the power of four.
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Weβre asked to find the derivative of a log function where the log is to the base four.
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Now we know that the derivative of the natural log of π₯ is one over π₯, where the natural log ln or ln π₯ is log to the base π of π₯.
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And since we know the derivative of the natural log of π₯, to find the derivative of a log which is not to the base π, we first convert our log to a natural logarithm.
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To do this, we use the base conversion identity.
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If we have a log to the base π of π’ which weβd like to convert to a logarithm of base π, then log to the base π of π’ is equal to log to the base π of π’ over log to the base π of π.
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In our case, π’ is equal to three π₯ squared plus two π₯ to the power of four.
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π is equal to four.
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And π, our new base, is equal to Eulerβs number π.
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Thatβs the base of a natural logarithm.
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So that by the base conversion identity, our log to the base four of three π₯ squared plus two π₯ to the power of four is equal to log to the base π of three π₯ squared plus two π₯ to the four over log to the base π of four.
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And using the notation that log to the base π of π₯ is ln or ln of π₯, we have π¦ equal to ln or ln of three π₯ squared plus two π₯ to the power of four over ln or ln four.
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Weβre asked to find dπ¦ by dπ₯.
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To do this, we first note that ln four is actually a constant.
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So the function weβre actually differentiating is log of three π₯ squared plus two π₯ to the four.
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And this is a function of a function.
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So weβre going to use the chain rule.
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This says that if π¦ is a function of π’, where π’ is a function of π₯, then dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯.
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In our case, we let π’ equal to three π₯ squared plus two π₯ to the four.
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And to find dπ’ by dπ₯, we use the power rule.
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This says that, for π not equal to negative one, the derivative of π times π₯ raised to the power π is π times π times π₯ raised to the power π minus one.
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That is, we multiply by the exponent and subtract one from the exponent so that dπ’ by dπ₯ is six π₯ plus eight π₯ to the power of three.
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With our substitution π’ is three π₯ squared plus two π₯ to the four, we have π¦ equal to one over the natural log of four times the natural log of π’.
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And as we saw earlier, the derivative of the natural log of π₯ is one over π₯ so that dπ¦ by dπ’ is one over the natural log of four times one over π’.
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And going back to our substitution π’ is equal to three π₯ squared plus two π₯ to the four, thatβs one over the natural log of four times one over three π₯ squared plus two π₯ to the four.
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Now we have everything we need to use the chain rule for dπ¦ by dπ₯. dπ¦ by dπ₯ is dπ¦ by dπ’ times dπ’ by dπ₯.
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And thatβs equal to one over the natural log of four times one over three π₯ squared plus two π₯ to the four times six π₯ plus eight π₯ cubed.
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We have a common factor of π₯ in both the numerator and the denominator, which gives us eight π₯ squared plus six over the natural log of four times two π₯ cubed plus three π₯.
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So dπ¦ by dπ₯ is equal to eight π₯ squared plus six over two π₯ cubed plus three π₯ times the natural log of four.