4.2.204 AP calculus practice question

[karush]

If $\displaystyle f(x)=\int_1^{x^3}\dfrac{1}{1+\ln t}\, dt$ for $x\ge 1$ then $f'(2)=$

(A) $\dfrac{1}{1+\ln 2}$

(B) $\dfrac{12}{1+\ln 2}$

(C) $\dfrac{1}{1+\ln 8}$

(D) $\dfrac{12}{1+\ln 8}$

ok I am little be baffled by this one due the $x^3$ in the limits
since from homework you just take integral and then plug in f'(x)

AP Calculus Exam PDF {WIP}
https://drive.google.com/file/d/131RCqemH8cHPh5FIW9VtqLPO4Kni0Bqo/view?usp=sharing

 

[MarkFL]

By the FTOC, we have:

\(\displaystyle f'(x)=\frac{d}{dx}\int_1^{x^3}\frac{1}{1+\ln(t)}\,dt=\frac{3x^2}{1+\ln(x^3)}\)

Hence:

\(\displaystyle f'(2)=\frac{12}{1+\ln(8)}\)

 

[karush]

well that's interesting
mahalo...

 

[I like Serena]

By the FTOC, we have:

\(\displaystyle f'(x)=\frac{d}{dx}\int_1^{x^3}\frac{1}{1+\ln(t)}\,dt=\frac{3x^2}{1+\ln(x)}\)

That is actually by the more advanced Leibniz integral rule.

And in this case $x^3$ should be substituted instead of $x$:

\(\displaystyle f'(x)=\frac{d}{dx}\int_1^{x^3}\frac{1}{1+\ln(t)}\,dt
= \left[\frac{1}{1+\ln(t)}\right]_{t=x^3}\cdot \d{}x(x^3) - \left[\frac{1}{1+\ln(t)}\right]_{t=1}\cdot\d{}x(1)
=\frac{3x^2}{1+\ln(x^3)}\)

 

[HallsofIvy]

In general, "Leibniz's rule": if [tex]f(x)= \int_{\alpha(x)}^{\beta(x)} g(x.t)dt[/tex] then
[tex]\frac{df(x)}{dx}= g(x, \beta(x))\frac{d\beta}{dx}- g(x,\alpha(x))\frac{d\alpha}{dx}+ \int_{\alpha(x)}^{\beta(X)} \frac{\partial g}{\partial x} dt[/tex].

That includes the basic "Fundamental theorem of Calculus" given in Calculus texts as a special case: if [tex]\beta(x)= x[/tex], [tex]\alpha(x)= a[/tex], a constant, and [tex]g(x,t)= g(t)[/tex] with no dependence on x, then [tex]\frac{d\beta}{dx}= 1[/tex], [tex]\frac{d\alpha}{dx}= 0[/tex], and [tex]\frac{\partial g}{\partial x}= 0[/tex] so
[tex]\left(\int_a^x g(t)dt\right)'= g(x)[/tex].

 

[MarkFL]

I've edited my post to correct it. I should have referred to my notes on the FTOC rather than "winging it." (Giggle)

 

[I like Serena]

I've edited my post to correct it. I should have referred to my notes on the FTOC rather than "winging it."

I consider the Leibniz integration rule difficult to remember and it's easy to make mistakes with it. (Giggle)
So I prefer to use 'just' FTOC and 'wing it'. (Nerd)

In this case, we can define:
$$G(u) = \int_1^u \frac{1}{1+\ln(t)}\,dt$$
Then:
$$f'(x)=\d{}x \int_1^{x^3}\frac{1}{1+\ln(t)}\,dt = \d{}x \Big(G(x^3) - G(1)\Big) = G'(x^3)\cdot 3x^2$$
Now we can apply FTOC to find:
$$G'(u) = \frac{1}{1+\ln(u)} \implies f'(x) = \frac{1}{1+\ln(x^3)}\cdot 3x^2$$

Like this we use only the chain rule and FTOC.
Less room for mistakes and no need to remember the Leibniz integration rule.