anyone
1) Differentiate
Y = x^2sinxtanx
f(x) = sinx ; f'(x) = cosx
g(x) = tanx ; g'(x) = sec^2
So I get:
Y' = 2cosxtanx + x^2sinxsec^2
2) Find an equation of the tangent line to the curve at the given point:
P(0,1)
Y = 1/(sinx + cosx)
Y' = 2sinx/(sinx + cosx)^2
this is obviously wrong based on the fact that you haven't answered the question. where is the tangent line? it is also wrong because you did not differentiate properly. use the quotient rule...or if it pleases your fancy, change the function to y = (sin(x) + cos(x))^{-1} and use the chain rule
we have three product here, you would have to take two at a time, so by the product rule you would say:
but for you would use the product rule again, to get, in total:
ALTERNATIVELY:
you could have realized that:
and use the quotient rule
you answer to the second was correct, but your answer is still incomplete. you were asked for the tangent line at a point
2) Find an equation of the tangent line to the curve at the given point:
P(0,1)
Y = 1/(sinx + cosx)
Y' = 2sinx/(sinx + cosx)^2
the derivative gives the formula formula for the slope at any value of x. plug in the required x-value into your derivative equation and that will give you the slope at that point. this will be the m in the equation of your line, y = mx + b
once you have m, you can use the point-slope form:
where is a point the line passes through, this is given to you. just solve for and you have your line