# Find dy/dx where y=e^(4xtanx)

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Here, we must apply both chain and product rules. The chain rule can be used to find the derivative of a function in the form ef(x), like this one. However it is useful to know that this will result in the following: f'(x)ef(x)... in other words the solution is always the derivative of the power times the initial equation. Knowing this can save a lot of time in the exam- it appears a lot.

Now, our only issue is finding the derivative of 4xtanx... this requires the product rule(the derivative of a product function uv= vdu+udv). In this example u=4x and v=tanx. Now du=4 and dv=sec2x. Slotting these into the above formula we get: 4tanx+4xsec2x.

All that is left is to bring together these two parts to get: d(e4xtanx)/dx= (4tanx+4xsec2x)e4xtanx.

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