Verified answer

f(x) = tan(2πx) - (√x).cos(πx)

according to the sum/difference rule, differentiate in parts, then add them together at the end

= d/dx tan(2πx) - d/dx (√x)*cos(πx)

# first part

d/dx tan(2πx)

break up in smaller functions, and differentiate them them, because you will need to do that anyway in the Rules

let h(x) = 2πx

and g(h(x)) = tan(h(x)) = tan(2πx),

by the chain rule

d/dx g(h(x)) = g'(h(x))*h'(x)

= sec²(2πx) * d/dx 2πx

= sec²(2πx) * 2π

= 2π*sec²(2πx)

notice: differentiate the outer function first, then write inner function in d/dx, then diffrentiate that. this approach simplifies differentials of more complex functions

## part 2:

d/dx (√x)*cos(πx)

let k(x) = √x

and n(x) = πx

and m(x) = cos(πx)

so that, d/dx k(x)m(x) = k'(x)m(x) + k(x)m'(n(x))

notice that m(n(x)) is more complex than k(x), make sure m'(n(x)) is at the end

d/dx k(x)m(n(x)) = k'(x)m(n(x)) + k(x)m'(n(x))

d/dx (√x)*cos(πx) = [1/(2√x)]*[cos(πx)] + [√x] * d/dx [cos(πx)]

now, d/dx m(n(x)) = m'(n(x)) * n'(x) = -sin(πx) * π

put this into the last term above

d/dx (√x)*cos(πx)

= [1/(2√x)]*[cos(πx)] + [√x] * [-sin(πx) * π]

= cos(πx)/(2√x) - π(√x)sin(πx)

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put them all together:

d/dx f(x)

= d/dx [g(h(x)) - k(x)m(n(x))]

= d/dx g(h(x)) - d/dx k(x)m(n(x))

= g'(h(x)) - k'(k)*m(n(x)) + k(x)*m'(n(x))*n'(x)

= 2π*sec²(2πx) - cos(πx)/(2√x) - π(√x)sin(πx)

{answer}

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hope this helps

be good

Correct

tan(2πx) - (√x)(cosπx)

d/dx tan(2πx) - (√x)(cosπx) =

= d/dx tan(2πx) - d/dx (√x)(cosπx) =

= d/dx sin(2πx)/cos(2πx) - d/dx (√x)(cosπx) =

= 2πcos(2πx)cos(2πx)-sin(2πx)[-2πsin(2πx]/cos(2πx)² - [(1/2√x * cos(πx) + √x * -π(sinπx)] =

= 2πcos(2πx)²+2πsin(2πx)² / cos(2πx)² - [cos(πx)/2√x -πsin(πx)√x]=

= 2π(cos(2πx)²+sin(2xπ)²) / cos(2πx)² - cos(πx)/2√x + πsin(πx)√x =

= 2π * 1/ cos(2πx)² -cos(πx)/2√x + πsin(πx)√x =

= 2πsec²(2πx)+ πsin(πx)√x -cos(πx)/2√x