How do you find the derivative of $u = (1 + e^{v}) (1 - e^{v})$ $u=(1+e^v)(1-e^v)$ ?

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How do you find the derivative of $u = (1 + e^{v}) (1 - e^{v})$ $u=(1+e^v)(1-e^v)$ ?

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Answer 1 · 2017-06-26T17:46:02

$\frac{d u}{d v} = - 2 e^{2 v}$ $(du)/(dv)=-2e^(2v)$

Explanation:

looking at the RHS we notice it is difference of squares when multiplied out, so it will be easier to differentiate if we multiply out first, rather than use the product rule then simplify.

$u = (1 + e^{v}) (1 - e^{v})$ $u=(1+e^v)(1-e^v)$

$\Rightarrow u = 1 - e^{2 v}$ $=>u=1-e^(2v)$

now differentiate wrt $v$ $v$

$\frac{d u}{d v} = 0 - 2 e^{2 v}$ $(du)/(dv)=0-2e^(2v)$

$\frac{d u}{d v} = - 2 e^{2 v}$ $(du)/(dv)=-2e^(2v)$

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How do you find the derivative of $u = (1 + e^{v}) (1 - e^{v})$ $u=(1+e^v)(1-e^v)$ ?

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