Are My Calculus Derivative Solutions Correct?

[alpha01]

Here are several problems (differentiation) i have attempted but not completely sure if they are correct

Homework Statement

differentiate y = sec(e^x)

The Attempt at a Solution

y = sec(e^x)

y = sec(u)

(where u = e^x)

dy/dx = tan(u)sec(u)e^x

tan(e^x)sec(e^x)e^x

If this is correct so far, can it be further factorized?:

e^x(tan sec(e^x))

Homework Statement

differentiate y = sec(log(x))

The Attempt at a Solution

y = sec(log(x))

y = sec(u)

(where u = log(x))

dy/dx = tan(u)sec(u) 1/x

= (tan sec(log(x))) / x

Homework Statement

find the second derivative of f(x) = sin(2x) + cos(3x)

The Attempt at a Solution

f(x) = sin(2x) + cos(3x)

f ' (x) = 2cos(2x) - 3sin(3x)

f '' (x) = cos(2x) - 4sin(2x) + sin(3x) - 9cos(3x)

 

[Mark44]

Here are several problems (differentiation) i have attempted but not completely sure if they are correct

Homework Statement

differentiate y = sec(e^x)

The Attempt at a Solution

y = sec(e^x)

y = sec(u)

(where u = e^x)

dy/dx = tan(u)sec(u)e^x

tan(e^x)sec(e^x)e^x

If this is correct so far, can it be further factorized?:

e^x(tan sec(e^x))

No. Your previous result is correct, but the simplification is not. The expression tan(e^x) sec(e^x) e^x is the product of its three factors. Your mistake was changing the product of the tangent and secant factors into a composition of them.

Homework Statement

differentiate y = sec(log(x))

The Attempt at a Solution

y = sec(log(x))

y = sec(u)

(where u = log(x))

dy/dx = tan(u)sec(u) 1/x

= (tan sec(log(x))) / x

Same comment as in the first question. Also, is the log function here the natural log? If so, you should use ln instead of log, which normally means log base 10.

Homework Statement

find the second derivative of f(x) = sin(2x) + cos(3x)

The Attempt at a Solution

f(x) = sin(2x) + cos(3x)

f ' (x) = 2cos(2x) - 3sin(3x)

f '' (x) = cos(2x) - 4sin(2x) + sin(3x) - 9cos(3x)

Your first derivative is correct, the the second derivative is not. You are apparently using the product rule on 2cos(2x) and 3sin(3x). You can use the product rule here, but it's more powerful than what you need, and it's easier to use the constant multiple rule (i.e., d/dx (k * f(x)) = k * d/dx(f(x)) = k*f'(x) ). Where you made your mistake was forgetting that d/dx(2) = 0 and d/dx(3) = 0 in the coefficients of the two terms in f'(x).
Mark