Can u guyz help me with these questions plz ??????

Question

1) How to Derivative these three questions?
a) y = 4^x + e^4x + π

b) y = 3x . ln (2x)

c) x^2 ÷ (sin x)^2

2) Simplify the expression as far as possible: ln(4) - 2ln(√ x) + ln(x/2)

Answer 1

1 a) The equation is y = 4^x + e^4x + π

Apply derivative to each side

d(y) = d(4^x + e^4x + π)

dy = d(4^x) + d(e^4x) + d(π)

Apply formulas: d(a^x) = a^x ln(a) and d(e^ax) = e^ax (a)

dy = 4^x ln(4) + e^4x (4) + 0

dy = 4^x ln(4) + 4 e^4x.

The solution is 4^x ln(4) + 4 e^4x.

Answer 2

1 b) The equation is y = 3x . ln (2x)

Apply derivative to each side

d(y) = d(3x × ln (2x))

Apply formula: d(uv) = u(dv) + v(du)

dy = 3x d(ln (2x)) + ln(2x)d(3x)

Apply formulas: d(ln(ax)) = 1/x and d(ax) = a

dy = 3x (1/x) + ln(2x)3

dy = 3 + 3ln(2x)

dy = 3{1 + ln(2x)}

The solution is dy = 3{1 + ln(2x)}.

Answer 3

C. Let y = x^2/(sinx)^2

Apply differentiation to each side

d(y) = d(x^2/(sinx)^2)

Apply formula: d(u/v) = (v*d(u)-u*d(v))/v^2

d(y) = ((sinx)^2*d(x^2)-x^2*d((sinx)^2))/((sinx)^2)^2

Apply formulas: d(x^n )= n*x^(n-1), d(sinx) = cosx

d(y) = ((sinx)^2*(2x)-x^2*2sinx*cosx)/(sinx)^4

d(y) = 2x(sinx)^2-2x^2sinxcosx / (sinx)^4

Taking sinx term common

d(y) = sinx(2xsinx-2x^2cosx)/(sinx)^4

d(y) = 2xsinx-2x^2cosx/(sinx)^3

Answer 4

2. Given log(4) - 2log(sqrtx) + log(x/2)

log(4) - 2log(x^(1/2)) + log(x/2)

Apply logarthimic formulas: alog(x)=logx^a and log(a/b) = loga-logb

log(4) - log(x^(1/2)*2) + logx - log(2)

log(4) - logx + logx - log(2)

log(4) - log(2)

Apply formula: loga - logb = log(a/b)

log(4/2)

log(2)

0.301

Answer 5

1 c) The expression is x^2 ÷ (sin x)^2

Apply derivative to each side

dy/dx = d(x^2 ÷ (sin x)^2)

Apply formula: d(u/v) = {v(du) - u(dv)}/ v^2

dy/dx = {(sinx)^2 d(x^2) - x^2 d(sinx)^2} / {(sinx)^2)}^2

dy/dx = {(sinx)^2 d(x^2) - x^2 d(sinx)^2} / {(sinx)^2)}^2

Apply formula: d(x^n) = n x^(n - 1)

dy/dx = {(sinx)^2 (2x)dx - x^2 (2 sinx(-cos x))dx} / {(sinx)^2)}^2

dy/dx = {(sinx)^2 (2x) - x^2 (2 sinx(-cos x))} / {(sinx)^2)}^2

dy/dx = {(sinx)^2 (2x) + 2x^2 (sinx(cos x))} / {(sinx)^2)}^2

The solution is 

dy/dx = {(sinx)^2 (2x) + 2x^2 (sinx(cos x))} / {(sinx)^2)}^.

 

Comments

y' = [2xsinx - 2x²cosx]/(sinx)³