Real Analysis fundamental theorems of Calculus contradiction? Evaluate: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt.$
Solution: $\frac{d}{dx} \int_{0}^{x} x^3t^3dt = \frac{7x^6}{4}.$
Proof:
Consider the function's $F:[a,b] \rightarrow \mathbb{R}$ and $f:[a,b] \rightarrow \mathbb{R}$, define: $F(t) = \frac{1}{4}x^3t^4 + c$ and $f(t) = x^3t^3.$ Observe that by proposition 4.7 F is differentiable on (a,b) and $F'(t) = x^3t^3 = f(t)$ for all x in $(a,b).$ Moreover by the polynomial property (proposition 2.17): F is continuous on [a,b] and f is continuous on [a,b], so by theorem 3.14 we know f is bounded and by theorem 6.18 f is integrable. We satisfy the criteria in the first fundamental theorem of calculus (theorem 6.22) hence:
$\frac{d}{dx} \int_{0}^{x} x^3t^3dt = \frac{d}{dx} [F(x) - F(0)] = \frac{d}{dx} \frac{1}{4}x^7 = \frac{7}{4}x^6.$
But by the second fundamental theorem of calculus (theorem 6.29):
$\frac{d}{dx} \int_{0}^{x} x^3t^3dt = f(x) = x^6 \neq F'(x) = \frac{7}{4}x^6$. Contradiction. WHY!?
(Textbook: Advanced Calculus - Patrick M. Fitzpatrick, Chapter 6.6)
Edit
Thank you everyone for your help, I managed to figure it out from all of the advice!
 A: When $f$ depends on $x$ and $t$ one has
\begin{equation}
\frac{d}{d x}\int_0^x f(x, t) d t = f(x, x) + \int_0^x\frac{\partial f}{\partial x}(x, t) d t
\end{equation}
In your case it gives
\begin{equation}
\frac{d }{d x}\int_0^x x^3 t^3 d t
= x^6 + \int_0^x 3 x^2 t^3 dt
\end{equation}
which is different from $x^6$.
A: It should be noted that the Leibniz Integral Rule deals with integrals like the one in question.
In the current example the integral takes the following form (with a slight generalization).
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= (x \, x)^n \, \frac{d}{dx}(x) - (x \cdot 0)^n \, \frac{d}{dx}(0) + \int_{0}^{x} \left[ \frac{d}{dx}(x \, t)^n \right] \, dt \\
&= x^{2n} + n \, x^{n-1} \, \int_{0}^{x} t^{n} \, dt \\
&= x^{2 n} + n \, x^{n-1} \cdot \frac{x^{n+1}}{n+1} \\
&= \frac{2 n + 1}{n+1} \, x^{2 n}.
\end{align}
A straight forward evaluation takes the form
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= \frac{d}{dx} \, \left( \frac{x^{2 n + 1}}{n+1} \right) = \frac{2n+1}{n+1} \, x^{2n}. 
\end{align}
By using a corollary of the Fundamental Theorem of Calculus, namely,
$$ \int_{a}^{b} f(t) \, dt = F(b) - F(a), $$
where $F(x)$ is the anti-derivative of $f(x)$, then
\begin{align}
\frac{d}{dx} \, \int_{0}^{x} (x \, t)^n \, dt &= \frac{d}{dx} \, \left[ x^n \, (F(x) - F(0) ) \right] \\
&= \frac{d}{dx} \, \left[ x^n \, \left(\frac{x^{n+1}}{n+1} - \frac{0^{n+1}}{n+1}\right) \right] \\
&= \frac{d}{dx} \, \left( \frac{x^{2n+1}}{n+1}\right) \\
&= \frac{2n+1}{n+1} \, x^{2n}.
\end{align}
Each of these views yields the same result.
