Determine the derivative for y=tanxcosx
a) y'=-sinxtanx + cosxsec^2x
b) y'=-sinxtanx + cosxtanx
c) y'=cosxsec^2x
d) y'=sinx - cosxsec^2x
Most of the above use the product rule,
(d/dx)u.v=u(dv/dx)+v(du/dx)
For example:
y=-sin(x)tan(x)
y'=-sin(x)d(tan(x))/dx + tan(x)d(-sin(x))/dx
=-sin(x)sec²(x)-tan(x)cos(x)
=-sin(x)/cos²(x)-sin(x)
=-sin(x)(1+sec²(x))
A is your answer because what you have there is the product rule[F'(x)*G(x)+G'(x)*F(x)].
To find the derivative of y = tan(x)cos(x), we can use the product rule.
The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by (u'v + uv').
Let's find the derivatives of the two functions:
u(x) = tan(x), so u'(x) = sec^2(x) [derivative of tan(x) is sec^2(x)]
v(x) = cos(x), so v'(x) = -sin(x) [derivative of cos(x) is -sin(x)]
Now, we can use the product rule:
y' = u'v + uv'
= sec^2(x)cos(x) + tan(x)(-sin(x))
= cos(x)sec^2(x) - sin(x)tan(x)
Therefore, the correct answer is:
a) y' = -sin(x)tan(x) + cos(x)sec^2(x)
To find the derivative of the function y = tan(x)cos(x), we will use the product rule and chain rule.
First, let's break down the function into two parts: u = tan(x) and v = cos(x).
The product rule states that if u and v are differentiable functions, then the derivative of their product uv with respect to x is given by the formula:
(d/dx)(uv) = u'v + uv'
Now, let's find the derivatives of u = tan(x) and v = cos(x).
Using the chain rule, the derivative of u = tan(x) can be found as follows:
u' = sec^2(x) * (d/dx)(x)
= sec^2(x)
Using the chain rule, the derivative of v = cos(x) can be found as follows:
v' = (d/dx)(cos(x))
= -sin(x)
Now, substitute the values we obtained into the product rule formula:
(d/dx)(tan(x)cos(x)) = (sec^2(x))(cos(x)) + (tan(x))*(-sin(x))
= cos(x)sec^2(x) - sin(x)tan(x)
Therefore, the correct derivative is y' = cos(x)sec^2(x) - sin(x)tan(x).
From the given options, the correct answer is a) y' = -sin(x)tan(x) + cos(x)sec^2(x).