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Answer 1 · 2018-03-11T11:48:40

$5 e^{4} + 2$ $5e^4+2$

First, we find the derivtaive of $y$ $y$ . Here, we must solve for $\frac{d}{d x} (e^{x^{2} + 3 x} + 2 x)$ $d/dx(e^(x^2+3x)+2x)$ .

Since $\frac{d}{d x} (f \pm g) = \frac{d}{d x} f \pm \frac{d}{d x} g$ $d/dx(f+-g)=d/dxf+-d/dxg$ , we write:

$\frac{d}{d x} e^{x^{2} + 3 x} + 2 \frac{d}{d x} x$ $d/dxe^(x^2+3x)+2d/dxx$

According to the chain rule, $(f(g(x)))'=f'(g(x))*g'(x)$

Here, $f(u)=e^u$ where $u=x^2+3x$ . We solve for:

$d/(du)e^u*d/dx(x^2+3x)$

$e^u(2x+3)$

Since $u=x^2+3x$ , we can say:

$e^(x^2+3x)(2x+3)$ , which we input into our earlier function:

$e^(x^2+3x)(2x+3)+2d/dxx$

$e^(x^2+3x)(2x+3)+2$

Now, we need to find the gradient at $x=1$ . The fact that the $y$ value is $e^4+2$ is irrelevant. Simply input that $x=1$ :

$e^(1^2+3*1)(2*1+3)+2$

$e^(1+3)(2+3)+2$

$5e^4+2$

The above is the gradient of $y$ when $x=1$ .