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(tanx)'=sec²x=1+tan²x #2

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lls3018 opened this issue Dec 14, 2017 · 0 comments

lls3018 commented Dec 14, 2017

i think this is right!

def backward(self, X, top_diff):
output = self.forward(X)
return (1.0 + np.square(output)) * top_diff

beacuse (tanx)'=sec²x=1+tan²x

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