Find f. f'(t) = sec(t)(sec(t) + tan(t)), = -1 2 (t) = tan(t) + sec(t) – 1- V2

Q: determine the general derivatives f(x)= -3 f(x)= sin x + cos x -sec2x

A: Simple use the identity of derivative of constant is zero and derivative of sinx is cosx and…

Q: Find_f'(x) 16. f(x) = sec² x - tan² x

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Q: f (x) = sec(x)(3x + √x) find the derivative

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Q: 2x2 tan z If f(x) = find sec z f (x) = Find f (5)=

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Q: tan (x) = sec (x)

A: We know that  tan(x) = sin(x)/cos(x)  Sec(x) = 1/cos(x)  So ,  sin(x)/cos(x) = 1/cos(x)

Q: y = sech /2 + v2x + 1 + tanh v8x² + 3

A: To find the derivative of the function

Q: Consider the function f(x) whose second derivative is ƒ''(x) = 9x + 6 sin(x). If ƒ(0) = 4 and f'(0)…

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Q: If f(t) = 2 - 9t cos t find f'(0).

A: Given :ft=2-9tcost

Q: 1 f(x) = x? cosx sin x + x' 4 3.

A: We will simply do the differentiation with respect to x for each term as follows

Q: f'(0) = sec0{ sec2 0 + tan2 0)

A: Assuming this expression is to simplify: Given: f'(0) =sec(0)( sec2 0 + tan2 0)

Q: Find f'(x) where f(x) sin(cos(4x2 +1))

A: Consider the given function

Q: Finf the equation of the tangent line to (cos(x)ex10x)/(Sec2(5x)+1)+11 at the point (0,11)

A: Given,           The equation of the tangent line to (cos(x)ex10x)(sec2(5x)+1)+11  at the point…

Q: Find the derivative

A: We are required to find the derivative of (a) y=cos2tan1+x3(b) f(t)=3csc5πt-71-cotπt5-7

Q: The derivative of y = 3cos(8x+6) is: 1. y' = 24 sin (8x+6) 2. y' = -24 sin (8x+6) 3. y' = -24 cos…

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Q: Solve the equation (y - (cos x)^2)dx + (cos x) dy = 0

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Q: The derivative of tan(9*x) is (sec(9*x))2(9))   Can you tell me where the sec2 came from?

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Q: Find the y'(0) when y = cos (sec²(2x))

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Q: Find f. f'(t) = sec(t)(sec(t) + tan(t)), = -1 f(t) =

A: Given that: f'(t)=sect(sect+tant),    -π2<t<π2, fπ4=-1

Q: Find the value of A B 3 16 fy' if y 314 3 16 In x =e4. when x = e X

A: according to bartleby we do only first part of multiple questions y=xlnxd dx(lnx)=1xd…

Q: Find f'(x) for f(x) = sin?x · tan (x² – 3x + 1)5).

A: We will find the derivative.