Changing the way you learn

Question	Answer
Express \(\sinh{x}\) in terms of exponentials.	\[\sinh{x}=\frac{e^{x}-e^{-x}}{2}\]
Express \(\cosh{x}\) in terms of exponentials.	\[\cosh{x}=\frac{e^{x}+e^{-x}}{2}\]
Express \(\tanh{x}\) in terms of exponentials.	\[\tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\] or \[\tanh x=\frac{e^{2x}-1}{e^{2x}+1}\]
Express \(\cosh^{-1}{x}\) in terms of natural logs	\[\cosh^{-1}{x}=\ln({x+\sqrt{x^{2}-1}})\]
Express \(\sinh^{-1}{x}\) in terms of natural logs	\[\sinh^{-1}{x}=\ln({x+\sqrt{x^{2}+1}})\]
Express \(\tanh^{-1}{x}\) in terms of natural logs	\[\tanh^{-1}{x}=\frac{\ln(\frac{1+x}{1-x})}{2}\]
Differentiate \(\cosh{x}\)	\[\frac{\partial \cosh{x}}{\partial x}=\sinh{x}\]
Differentiate \(\sinh{x}\)	\[\frac{\partial \sinh{x}}{\partial x}=\cosh{x}\]
Differentiate \(\tanh{x}\)	\[\frac{\partial \tanh{x}}{\partial x}=\sech^{2}{x}\]
Differentiate \(\sinh^{-1}{x}\)	\[\frac{\partial sinh^{-1}{x}}{\partial x}=\frac{1}{\sqrt{1+{x}^2}}\]
Differentiate \(\cosh^{-1}{x}\)	\[\frac{\partial cosh^{-1}{x}}{\partial x}=\frac{1}{\sqrt{{x}^{2}-1}}\]
Differentiate \(\tanh^{-1}{x}\)	\[\frac{\partial tanh^{-1}{x}}{\partial x}=\frac{1}{1-{x}^{2}}\]
Integrate \(\tanh^{-1}{x}\)	\[\int \tanh x = \ln{\cosh{x}}\]
What is Osborne's Rule?	The idea that trigonometric equations can be changed to hyperbolic equations by exchanging the trig functions for their hyperbolic counterparts and changing the sign wherever there is a product of two sines.
Use Osborne's rule to find the corresponding hyperbolic equation to: \[\tan^{2}{x}+1\equiv \sec^{2}{x}\]	\[1=sech^{2}{x} + \tanh^{2}{x}\]
Use Osborne's rule to find the corresponding hyperbolic equation to: \[\cot^{2}{x}+1\equiv \cosec^{2}{x}\]	\[cosech^{2}{x}=1+ \coth^{2}{x}\]
What is the integral of\( f'(x)[f(x)]^{n}\)	\[\frac{1}{n+1}[f(x)]^{n+1}+c\]
What is the integral of \(\frac{f'(x)}{f(x)}\)	\[\ln\left \| f(x) \right \|+c\]

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WJEC FP3

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	Created by Joshua Butterwor about 8 years ago