Need help with a derivative:

$$\frac{ d }{ dt } e ^{kcos(2t)}$$

I think this should be : $$ke ^{kcos(2t)} *\frac{ d }{ dt } kcos(2t)$$

Which is then : $$ke ^{kcos(2t)} * -ksin(2t)$$

Question

Need help with a derivative: 

$$\frac{ d }{ dt } e ^{kcos(2t)}$$

I think this should be : $$ke ^{kcos(2t)} *\frac{ d }{ dt } kcos(2t)$$

Which is then : $$ke ^{kcos(2t)} * -ksin(2t)$$

anonymous · Answer

but apparently this is wrong. Does anyone see where I went wrong?

vishweshshrimali5 · Answer

You did several mistakes in differentiating the given function.

vishweshshrimali5 · Answer

First of all while differentiating the first time you did not follow the chain rule properly.

anonymous · Answer

oh ya, i should have brough the whole thing down, not just a k....

vishweshshrimali5 · Answer

It should have been:

$$\large{\cfrac{ d }{ dt } e ^{kcos(2t)}}$$

$$\large{=e^{k\cos(2t)} * \cfrac{d}{dt}(k\cos(2t))}$$

anonymous · Answer

oh

vishweshshrimali5 · Answer

Now, while calculating derivative of kcos(2t), you did the same mistake. You differentiated cos correctly but forgot to differentiate the angle i.e. (2t).

vishweshshrimali5 · Answer

So, it should have been:

$$\large{\cfrac{d}{dt}(k\cos(2t))}$$

$$\large{= k*(-\sin(2t))*2}$$

$$\large{=-2k\sin(2t)}$$

vishweshshrimali5 · Answer

Did you get it ?

anonymous · Answer

ya, i didnt realize the chain rule applied to the angle as well...

vishweshshrimali5 · Answer

Its okay.

anonymous · Answer

thanks for clearing that up, its been quite a while :D

vishweshshrimali5 · Answer

It was my pleasure to be of some help.

Best of luck for further questions and good night (according to Indian Standard Time)!!!