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!

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.
