# Differentiation questions (alot)

Q1)

Q2)

Q3)

Q4)

Q5)

Q6)

asked Aug 10, 2014 in CALCULUS
edited Aug 10, 2014

Q1)

a) f(x)=sin(6x)

Take derivative with respect to x

f'(x)=d/dx(sin(6x))

=cos 6x d/dx(6x)

=6cos 6x

b) f(x)=cos(3x^2+x-1)

Take derivative with respect to x

f'(x)=d/dx(cos(3x^2+x-1))

=-sin(3x^2+x-1) d/dx(3x^2+x-1)

=- (6x+1)sin(3x^2+x-1)

c) f(x)=tan(-4x+7)

Take derivative with respect to x

f'(x)=d/dx(tan(-4x+7))

=sec^2(-4x+7) d/dx(-4x+7)

=-4sec^2(-4x+7)

d) f(x)=sec(cosx)

Take derivative with respect to x

f'(x)=d/dx(sec(cosx))

=sec(cosx)tan(cosx) d/dx(cosx)

=sec(cosx)tan(cosx)(-sinx)

=-sec(cosx)tan(cosx)sinx

e) f(x)=cotxsinx

Take derivative with respect to x

f'(x)=d/dx(cotxsinx)

Use product rule:d/dx(uv)=uv'+vu'

v=sinx    v' =cosx

u = cotx     u'= - cosec^x

f'(x)=cotxd/dx(sinx)+sinxd/dx(cotx)

=cotx(cosx) +sinx(-cosec^2x)

=(cosx/sinx)(cosx) +sinx(-1/sin^2x)

=cos^2x/sinx-1/sinx

=(cos^2x-1)/sinx

=-(-cos^2x+1)/sinx

=(sin^2x)/sinx

=sinx

edited Aug 11, 2014 by bradely

f) f(x)=x^5cosecx

Take derivative with respect to x

f'(x)=d/dx(x^5cosecx)

Use product rule:d/dx(uv)=uv'+vu'

u=x^5    u' =5x^4

v = cosecx     v'= - cosecxcotx

f'(x)=x^5 d/dx(cosecx)+cosecxd/dx(x^5)

=-x^5cosecxcotx +cosecx(5x^4)

= -x^5cosecxcotx +5x^4cosecx

g) f(x)=e^cosecx

Take derivative with respect to x

f'(x)=d/dx(e^cosecx)

f'(x)=e^cosecx d/dx(cosecx)

=-e^cosecx cosecxcotx

h) f(x)=x^6e^x

Take derivative with respect to x

f'(x)=d/dx(x^6e^x)

Use product rule:d/dx(uv)=uv'+vu'

u=x^6    u' =6x^5

v =e^x     v'= e^x

f'(x)=x^6 d/dx(e^x )+e^x d/dx(x^6)

=x^6e^x +e^x(6x^5)

=x^5e^x (x+6)

=x^5 e^x(x+6)

i) f(x)=x^5lnx

Take derivative with respect to x

f'(x)=d/dx(x^5lnx)

Use product rule:d/dx(uv)=uv'+vu'

u=x^5    u' =5x^4

v =lnx     v'= 1/x

f'(x)=x^5 d/dx(lnx)+cosecxd/dx(x^5)

=x^5(1/x) +lnx(5x^4)

=x^4 +lnx(5x^4)

= x^4(1+5lnx)

j) f(x)=e^x/(6+lnx)

Take derivative with respect to x

f'(x)=d/dx(e^x/(6+lnx))

Use Quotient rule:d/dx(u/v)=(vu'-uv')/v^2

u=e^x    u' =e^x

v = 6+lnx  v'=1/x

f'(x)=((6+lnx)d/dx(e^x)-e^xd/dx(6+lnx))/(6+lnx)^2

=((6+lnx)(e^x)-e^x(1/x))/(6+lnx)^2

=e^x(6+lnx-1/x)/(6+lnx)^2

Q2)

f(x) =(sinx+1)/(secx-1)

Take derivative with respect to x

f'(x)=d/dx((sinx+1)/(secx-1))

Use Quotient rule:d/dx(u/v)=(vu'-uv')/v^2

u=sinx+1    u' =cosx

v = secx-1  v'=secxtanx

f'(x)=((secx-1)(cosx)-(sinx+1)(secxtanx))/(secx-1)^2

=((1-secx)-(sinxsecxtanx+secxtanx)/(secx-1)^2

=((1-secx)-(tan^2x+secxtanx)/(secx-1)^2

Q3)

f(t) =tcostcott

Take derivative with respect to t

f'(x)=d/dt(tcostcott)

=td/dt(costcott)+costcottd/dt(t)

=t[costd/dt(cott)+cottd/dt(cost)]+costcott

=t[cost(cosec^2t)+cott(-sint)]+costcott

=t[cost(1/sin^2t)-cost/sint(sint)]+costcott

=t[cottcosect-cost]+costcott

Q4)

f(y) =√(1-2tany)

Take derivative with respect to y

f'(y)=d/dy(√(1-2tany))

Use formula:d/dx(√x)=1/2√x

f'(y)=(1/2√(1-2tany))d/dy(1-2tany)

=(1/2√(1-2tany))(-2sec^2y)

=(1/√(1-2tany))(-sec^2y)

= (-sec^2y/√(1-2tany))

Q5)

a)

f(x)=6(e)^x5

Take derivative with respect to x

f'(x)=d/dx(6(e)^x5 )

=6(e)^x5 d/dx(x5)

=30x^4(e)^x5

Again take derivative both sides

f''(x)=d/dx(-3x²sin x³)

Use product rule:d/dx(uv)=uv'+vu'

u=x²          v' =2x

v =sin x³     u'= 3x^2cos x³

f''(x)=-3[x²d/dx(sin x³)+sin x³d/dx(x²)]

=-3[x²(3x²cos x³)+sin x³(2x)]

=-3[3x4cos x³+2xsin x³]

----------------------------------

Q6)

f(x)=t^6ln(5t)

Take derivative with respect to t

f'(x)=d/dt(t^6ln5t)

Use product rule:d/dx(uv)=uv'+vu'

u=t^6    u' =6t^5

v =ln5t     v'= 5/t

f'(x)=t^6 d/dx(ln5t)+ln(5t)d/dt(t^6)

=t^6(5/t) +ln5t(6t^5)

=t^5(5) +ln5t(6t^5)

= x^5(5+6ln5t)