Question

What is the derivative of \[\cos ecx\] ?

Answer 1

Hint: We already know the formula for finding the derivative of \[\cos ecx\] . But we will derive the answer properly using formulas and relations between the trigonometric identities.
Formula used:
Quotient rule \[\dfrac{d}{{dx}}\left[ {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \dfrac{{g\left( x \right)f'\left( x \right) - f\left( x \right)g'\left( x \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\]

Complete step-by-step answer:
We know that,
derivative of \[\cos ecx\] is written or expressed as,
 \[\dfrac{d}{{dx}}\cos ecx\]
Now we will start deriving the formula,
We know that, \[\cos ecx = \dfrac{1}{{\sin x}}\]
Thus we can write,
 \[\dfrac{d}{{dx}}\dfrac{1}{{\sin x}}\]
Now we will use the quotient rule,
 \[ = \dfrac{{\sin x\left( 0 \right) - \cos x \times 1}}{{{{\left( {\sin x} \right)}^2}}}\]
On solving the terms we get,
 \[ = \dfrac{{ - \cos x}}{{{{\left( {\sin x} \right)}^2}}}\]
Now we will separate the terms as,
 \[ = \dfrac{{ - 1}}{{\sin x}}.\dfrac{{\cos x}}{{\sin x}}\]
We know that reciprocal of sin function is cosec and the ratio of cos to sin function is cot function. thus,
 \[ = - \cos ecx.\cot x\]
Thus we get the solution as,
 \[\dfrac{d}{{dx}}\cos ecx = - \cos ecx.\cot x\]
So, the correct answer is “ \[\dfrac{d}{{dx}}\cos ecx = - \cos ecx.\cot x\] ”.

Note: Note that the way we write the answer depends on the marks it is assigned for. Also note that sometimes in multiple choice questions they may give the second last step of the solution above as the derivative of \[\cos ecx\] but we as usual make it false. There we make mistakes. How?
 \[\dfrac{{ - 1}}{{\sin x}}.\dfrac{{\cos x}}{{\sin x}} = - \cos ecx.\cot x\]
This is the same answer but only more simplification makes it change. So do carefully read the answer. This frequently happens in trigonometry related questions. So one should be clear with the relations of the ratios.