Asked by gnozahs
find the derivative of f(x)=tanx-4/secx
I used the quotient rule and got (sec^2x-4)(secx)-(tanx-4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks.
I used the quotient rule and got (sec^2x-4)(secx)-(tanx-4)(cosx)/sec^2x but I'm pretty sure that's wrong. Help. Thanks.
Answers
Answered by
MathMate
It is not clear what your expression really is, there are two possibilities. In both cases, it pays to do a little simplification before differentiating.
If f(x)=(tan(x)-4)/sec(x)
then
f(x)
=tan(x)*cos(x) - 4 cos(x)
=sin(x) - 4 cos(x)
f'(x) = cos(x) + 4 sin(x)
If f(x) = tan(x) - 4/sec(x)
then
f(x) = tan(x) - 4 cos(x)
f'(x) = sec<sup>2</sup>(x) + 4 sin(x)
If f(x)=(tan(x)-4)/sec(x)
then
f(x)
=tan(x)*cos(x) - 4 cos(x)
=sin(x) - 4 cos(x)
f'(x) = cos(x) + 4 sin(x)
If f(x) = tan(x) - 4/sec(x)
then
f(x) = tan(x) - 4 cos(x)
f'(x) = sec<sup>2</sup>(x) + 4 sin(x)
Answered by
gnozahs
cos(x) + 4 sin(x) is correct. How did the 4 end up with the sin and not with the cos?
Answered by
MathMate
It comes from
-4/sec(x) = -4 cos(x)
So when differentiated, it gives 4 sin(x)
-4/sec(x) = -4 cos(x)
So when differentiated, it gives 4 sin(x)
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