For the function is defined by
On which two intervals is the function increasing

I got -11 to 6 as one interval

and only part of the other one which is (some # to 10)

2. what did you get for f'(x) ?

3. (7x^6)(x+6)^4 + 4(x+6)^3(x^7)

4. (7x^6)(x+6)^4 + 4(x+6)^3(x^7)
fine, but did you simplify it before you set it equal to 0?

factoring out what the two terms have in common will help alot.

5. does it become [x^6(x+6)^3] 7(x+6) + 4x

6. cleaned up your derivative a little bit ...
$f'(x) = 7x^6(x+6)^4 + 4x^7(x+6)^3$

now, look at the following expression ... what can you factor out of those two terms (in other words, what factors do they have in common to "pull" out?)

$7x^6y^4 + 4x^7y^3$

7. Hello, Lambo!

For $x \in [-11, 10]$, the function $f(x)$ is defined by: . $f(x) \:=\:x^7(x+6)^4$
On which two intervals is the function increasing ?

I got -11 to 6 as one interval . . . . sorry, no
The function is increasing when $f'(x)$ is positive.

The derivative is: . $f'(x) \;=\;x^7\cdot4(x+6)^3 + 7x^6\cdot(x+6)^4 \;=\;x^6(x+6)^3\bigg[4x + 7(x+6)\bigg]$

So we have: . $f'(x)\;=\;x^6\cdot(x+6)^3\cdot(11x+42)\;>\;0$

This product is positive when: . $\begin{Bmatrix}(1) & \text{all three factors are positive} \\ (2) & \text{two factors are negative} \end{Bmatrix}$

(1) All three factors are positive.

. . $x^6$ is always positive, except when $x = 0$

. . $(x+6)^3 \:>\:0\quad\Rightarrow\quad x \:>\:\text{-}6$

. . $11x+42 \:>\:0 \quad\Rightarrow\quad x \:>\:\text{-}\tfrac{42}{11}$

All three are satisfied if: . $x \:>\:\text{-}\frac{42}{11}\:\text{ and }\:x \neq 0$

. . The interval is: . $\boxed{\;\left(\text{-}\tfrac{42}{11},\:0\right) \cup (0,\:\infty)\;}$

(2) Two factors are negative.

. . $x^6$ is always positive, except when $x = 0$

. . $(x+6)^3 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}6$

. . $11x+42 \:<\:0 \quad\Rightarrow\quad x \:<\:\text{-}\tfrac{42}{11}$

All three are satisfied if: . $x \:<\:\text{-}6$

. . The interval is: . $\boxed{\;(\text{-}\infty,\:\text{-}6)\;}$