Find the equation of the tangent line to the graph of $e^{x^7-8x}$ at $(-1,e^7)$

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Find the equation of the tangent line to the graph of $g(x)=e^{x^7-8x}$ at $(-1,e^7)$


$Solution:$

The value of the slope of the tangent line to $g(x)$ at the point $(-1,e^7)$ is $g'(-1)$, so lets calculate that.

$g'(x)=e^{x^7-8x}(x^7-8x)'=e^{x^7-8x}(7x^6-8)$

$\rightarrow g'(-1)=e^{(-1)^7-8(-1)}(7(-1)^6-8)=e^{7}(-1)=-e^7$

So we know that $y=-e^7x+b$. Let's plug in the given point to solve for $b$.

$e^7=-e^7(-1)+b$

$\rightarrow b=0$

Thus our equation is $y=-e^7x$

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Well done! Your equation is correct.