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Derivative of e^2x
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2e^2x
can u explain how u get that is it just using the power rule with the 2
\[\frac{d}{dx}e^{kx} = ke^{kx}\]
\[\LARGE (e^u)'=e^u\cdot u'\]
have you done the chain rule yet?
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yea
so we differentiate e^u and then differentiate u and multiply them together in this case u = kx so du/dx = k
i dont get it
which bit confuses you, the chain rule bit?
yea
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ok ill explain it better than i did
is this how you learnt it? \[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]
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