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I've spent almost 6 total hours hacking at this problem. And I always end up by a factor of 3 in one of the terms when checked against Wolfram's derivative calculator, which is correct when I manually calculate the derivative directly from the source equation in excel.

I'd love to have someone show me where I'm making my error.

Here we go...

y = 3t(5t + 4)^5 ln(y) = ln(3t(5t + 4)^5) ln(y) = 5*ln(3t(5t + 4)) d/dt(ln(y)) = d/dt(5*ln(3t(5t + 4))) d/dt(ln(y)) = d/dt(5)*d/dt(ln(3t(5t + 4))) d/dt(ln(y)) = d/dt(ln(3t(5t + 4))) d/dt(ln(y)) = 1/(3t(5t + 4)) *d/dt(3t(5t + 4)) d/dt(ln(y))*(3t(5t + 4)) = d/dt(3t(5t + 4)) d/dt(ln(y))*(3t(5t + 4)) = d/dt(15t^2 + 12t) d/dt(ln(y))*(3t(5t + 4)) = d/dt(15t^2) + d/dt(12t) d/dt(ln(y))*(3t(5t + 4)) = 15*d/dt(t^2) + 12*dt/dt(t) d/dt(ln(y))*(3t(5t + 4)) = 15*d/dt(t^2) + 12 d/dt(ln(y))*(3t(5t + 4)) = 15*2t*dt/dt + 12 d/dt(ln(y))*(3t(5t + 4)) = 15*2t + 12 d/dt(ln(y))*(3t(5t + 4)) = 30t + 12 d/dt(ln(y)) = (30t + 12)/(3t(5t + 4)) 1/y*d/dt = (30t + 12)/(3t(5t + 4)) d/dt = y*(30t + 12)/(3t(5t + 4)) d/dt = 3t(5t + 4)^5 *(30t + 12)/(3t(5t + 4)) d/dt = (5t + 4)^4 *(30t + 12) d/dt = [6(5t + 4)^4 *(5t + 2)]

Wolfram:

d/dt = 6(5t + 4)^4 *(15t + 2)

Amended approach as per nbubis:

The log method was presented as an easy way to handle complex exponents. So I adopted it for all exponent handling.

y = 3t(5t + 4)^5 d/dt(y) = d/dt(3t(5t + 4)^5) F(t) = f(t)*g(h(t)) f(t) = 3t f'(t) = 3 g(t) = (t)^5 g'(t) = 5(t)^4 h(t) = 5t + 4 h'(t) = 5 F'(t) = f'(t)*g(h(t)) + f(t)*g'(h(t)) F'(t) = f'(t)*g(h(t)) + f(t)*g'(h(t))*h'(t) F'(t) = 3*(5t + 4)^5 + 3t*5(5t + 4)^4*5 F'(t) = 3*(5t + 4)^5 + 75t(5t + 4)^4

How did you remove ^5?

Duh, common factor!!!

F'(t) = (3*(5t + 4) + 75t)(5t + 4)^4 F'(t) = 3(5t + 4 + 25t)(5t + 4)^4 F'(t) = 3(30t + 4)(5t + 4)^4 F'(t) = 3*2(15t + 2)(5t + 4)^4 F'(t) = [6(15t + 2)(5t + 4)^4]

Thanks!

And now for the complete log method:

y = 3t(5t + 4)^5 ln(y) = ln(3t(5t + 4)^5) ln(y) = ln(3) + ln(t) + ln((5t + 4)^5) ln(y) = ln(3) + ln(t) + 5*ln(5t + 4) d/dt(ln(y)) = d/dt(ln(3)) + d/dt(ln(t)) + d/dt(5*ln(5t + 4)) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5*d/dt(ln(5t + 4)) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*d/dt(5t + 4) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*(d/dt(5t) + d/dt(4)) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*(d/dt(5t) + 0) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*(d/dt(5t)) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*(5*dt/dt) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 5/(5t + 4)*(5) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*d/dt(t) + 25/(5t + 4) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t*dt/dt + 25/(5t + 4) d/dt(ln(y)) = 1/3*d/dt(3) + 1/t + 25/(5t + 4) d/dt(ln(y))3 = d/dt(3) + 3/t + 3*25/(5t + 4) d/dt(ln(y))3 = 0 + 3/t + 75/(5t + 4) d/dt(ln(y))3 = 3/t + 75/(5t + 4) d/dt(ln(y))3t = 3 + 75t/(5t + 4) d/dt(ln(y))3t(5t + 4) = 3(5t + 4) + 75t d/dt(ln(y)) = (3(5t + 4) + 75t)/3t(5t + 4) 1/y*dy/dt = (3(5t + 4) + 75t)/3t(5t + 4) dy/dt = y(3(5t + 4) + 75t)/3t(5t + 4) dy/dt = 3t(5t + 4)^5 *(3(5t + 4) + 75t)/3t(5t + 4) dy/dt = (5t + 4)^4 *(3(5t + 4) + 75t) dy/dt = (5t + 4)^4 *(15t + 12 + 75t) dy/dt = (5t + 4)^4 *(90t + 12) dy/dt = [6(5t + 4)^4 *(15t + 2)]!!!

This has been very educational.

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    $\begingroup$ Errors start at the beginning: We want $\ln(3t)+5\ln(5t+4)$. Or better $\ln 3+\ln t+5\ln(5t+4)$. $\endgroup$ Commented Jul 4, 2014 at 21:31
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    $\begingroup$ Am I missing something, why so many down votes on perfectly reasonable answers? Ok this is not easiest way to differentiate the above, but all the answers adhere to solving the OPs question and not trying to do a completely different approach..+1 to all! $\endgroup$ Commented Jul 4, 2014 at 23:43

4 Answers 4

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In the third line you go from $$\ln y = \ln 3t(5t+4)^5$$ to $$\ln y = 5 \ln 3t(5t+4)$$ but you can't extract the power $5$ at this stage because you don't have $(3t)^5$ inside the logarithm. Instead note (as in Andre Nicholas comment) that $$\ln 3t(5t+4)^5=\ln 3+\ln t+5\ln (5t+4)$$ and the factor $3$ drops out as a constant.

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Why take the logarithm? just do the straight forward calculation: $$y = 3t(5t + 4)^5$$ By the product rule: $$\frac{dy}{dt} = 3(5t + 4)^5 + 3t \cdot 25(5t+4)^4 $$ $$=\left(3(5t+4)+75t\right)(5t+4)^4$$ $$=6(15t+2)(5t+4)^4$$

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  • $\begingroup$ @redthumb - $(5t+4)^5 = (5t+4)(5t+4)^4$. $\endgroup$ Commented Jul 4, 2014 at 22:31
  • $\begingroup$ I'm referring to the product rule expansion where 3(5t + 4)^5 is reduced to 3(5t + 4)? See my amended original above. Thanks! $\endgroup$ Commented Jul 4, 2014 at 22:39
  • $\begingroup$ @redthumb - My comment shows exactly how you remove the $^5$ - it just becomes a power of $4$ with $(5t+4)$ factored out. $\endgroup$ Commented Jul 4, 2014 at 23:47
  • $\begingroup$ Got it, common factor. Can't believe I didn't see it. Thanks again! $\endgroup$ Commented Jul 5, 2014 at 0:15
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In your third line, you applied the rule of the exponent prematurely. That exponent $5$ applies only to the second factor in the argument of log.

First, undo the product

$$\ln y = \ln(3t(5t + 4)^5 ) \iff \ln y = \ln (3t) + \ln((5t+4)^5) $$

Now you can write $$\ln y = \ln (3t) + 5\ln((5t+4)) = \ln 3 + \ln(t) + 5\ln(5t+4) $$

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The following step is incorrect:

d/dt(ln(y)) = d/dt(5*ln(3t(5t + 4)))

d/dt(ln(y)) = d/dt(5)*d/dt(ln(3t(5t + 4)))

It is not true that $(d/dt)\Big( AB \Big)= \Big((d/dt)A\Big)\Big((d/dt)B\Big)$. In particular, it is not true that $(d/dt)(5A) = \Big((d/dt)5\Big)\Big((d/dt) A\Big).$

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