Derivative of $e^{2x^4-x^2-1}$ with limit definition of derivative Let $f:\mathbb{R}\to\mathbb{R}$ be defined as
$f(x) = e^{2x^4-x^2-1}$.
I have to find the derivative using the defintion:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
My approach:
$$
\begin{align}
&\lim_{h \to 0} \frac{\exp({2(x+h)^4-(x+h)^2-1})-\exp({2x^4-x^2-1})}{h}\\
&=\lim_{h \to 0} \frac{\exp(2 h^4 + 8 h^3 x + h^2 (12 x^2 - 1) + h (8 x^3 - 2 x) + 2 x^4 - x^2 - 1) - e^{2 x^4 - x^2 - 1}}{h}
\end{align}$$
How can I remove $h $from denominator?

Hint
Factor the term $e^{2 x^4 - x^2 - 1}$ in each of the two terms of the difference. Then you get the product
$$\frac{f(x+h)-f(x)}{h}=e^{2 x^4 - x^2 - 1} \times \frac{e^{hA(x,h)}-1}{h}$$
Where $A(x,h)=h(8x^2 -x) +h^2 B(x,h)$ with $A,B$ polynomials.
Then as $e^{y} -1 \approx y$ around $0$, you get
$$e^{2 x^4 - x^2 - 1} \times \frac{e^{hA(x,h)}-1}{h} \approx e^{2 x^4 - x^2 - 1} \left((8x^2-x) +hB(x,h)\right)$$
which converges to $(8x^2-x)e^{2 x^4 - x^2 - 1}$ as $h \to 0$. Proving that $f^\prime(x)=(8x^2-x)e^{2 x^4 - x^2 - 1}$.
