How do you differentiate #f(t)=tan(e^t)+e^(tant)#?

Question

Christopher Allers · Accepted Answer

Use the chain rule.

Using the chain rule, we first take the derivative of the "outside" term, then multiply the result by the derivative of the "inside" term. For the first half of the function, #tan(e^t)#, we take the derivative of tan, which is #sec^2#, and we leave the inside term (#e^t#) alone. This gives us #sec^2(e^t)#. Now we multiply this derivative by the derivative of the "inside" term, which is still just #e^t#. For the first half of the derivative, we have #e^t*sec^2(e^t)#. For the second half of the function, we take the derivative of #e#, which is still just #e#. This gives us, still, #e^tan(t)#. Now we take the derivative of the "inside" term, #tan(t)#, which is #sec^2(t)#. These multiply, just as above, to give the second half of the derivative as #e^(tan(t))*sec^2(t)#. We now simply add the derivatives, as specified by our original function. Our final answer is: #f'(t)=e^tsec^2(e^t)+e^(tan(t))sec^2(t)#

Elijah Abram · Answer

undefined To differentiate $ f(t) = \tan(e^t) + e^{\tan(t)} $, apply the chain rule and the derivative of the tangent function.

The derivative of $ \tan(x) $ is $ \sec^2(x) $.

The derivative of $ e^x $ is $ e^x $.

Using the chain rule, the derivative of $ \tan(e^t) $ is $ \sec^2(e^t) \cdot e^t $.

The derivative of $ e^{\tan(t)} $ is $ e^{\tan(t)} \cdot \sec^2(t) $.

Therefore, the derivative of $ f(t) $ is $ \frac{d}{dt}( \tan(e^t) + e^{\tan(t)}) = \sec^2(e^t) \cdot e^t + e^{\tan(t)} \cdot \sec^2(t) $.