Derivative of (4 secx tanx)
y = 4secx tanx
by the product rule ....
dy/dx = 4secx(sec^2 x) + 4tanx(secxtanx)
= 4sec^3 x + 4secx tan^2x
= 4secx(sec^2 x + tan^2 x) , but sec^2 x - 1 = tan^2 x
= 4secx ( sec^2 x + sec^2x + 1)
= 4secx(2sec^2 x + 1) or 8sec^3 x + 4secx
d/dx [ 4 * 1/cos x * sin x /cos x ]
= 4 d/dx [ sin x/cos^2 x}
=4 [ cos^3 x - sin x(2 cos x * -sin x) ] / cos^4 x
=4 [ cos^3 x - sin 2x + sin^2 x ] / cos^4 x
4 [ cos^3 x - sin x(2 cos x * -sin x) ] / cos^4 x
= (4cos^2x + 2sin^2x)/cos^3x
= (2cos^2x+2)/cos^3x
To find the derivative of the expression (4 sec(x) tan(x)), you can use the product rule and the chain rule.
Let's break down the expression first:
f(x) = 4 sec(x) tan(x)
The product rule states that if you have two functions, u(x) and v(x), the derivative of their product is given by:
d/dx (u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = 4 sec(x) and v(x) = tan(x). Now we can find the derivatives of each function:
u'(x) = d/dx (4 sec(x))
v'(x) = d/dx (tan(x))
To find the derivative of u(x) = 4 sec(x), we can rewrite it as:
u(x) = 4 * (1/cos(x))
Now we can differentiate u(x) using the chain rule, which states that if you have a function g(x) inside another function f(x), the derivative is given by:
d/dx (f(g(x))) = f'(g(x)) * g'(x)
In this case, f(u) = 4 * u and g(x) = 1/cos(x). The derivatives are:
f'(u) = 4
g'(x) = d/dx (1/cos(x))
To find g'(x), we differentiate 1/cos(x):
g'(x) = -sin(x)/cos^2(x)
Now we can substitute these values back into the product rule equation:
d/dx (4 sec(x) tan(x)) = u'(x) * v(x) + u(x) * v'(x)
= (4 * (1/cos(x))) * tan(x) + (4 sec(x)) * (-sin(x)/cos^2(x))
Simplifying this expression, we get:
d/dx (4 sec(x) tan(x)) = (4 * tan(x))/cos(x) - (4 sec(x) * sin(x))/cos^2(x)
Therefore, the derivative of (4 sec(x) tan(x)) is: (4 * tan(x))/cos(x) - (4 sec(x) * sin(x))/cos^2(x).