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Math`x^4(x+y)=y^2(3xy)` `x^5+x^4y=3xy^2y^3` Differentiating with respect to x.We get `5x^4+(4x^3y+x^4(dy/dx))=(3y^2+3x(2y)(dy/dx))3y^2(dy/dx)` `(5x^4+4x^3y3y^2)=(6xy3y^2x^4)(dy/dx)`...

MathNote: 1) If y = e^x ; then dy/dx = e^x 2) If y = x^n ; then dy/dx = n*x^(n1) ; where 'n' = real number 3) If y = u*v ; where both u & v are functions of 'x' ; then dy/dx = u*(dv/dx) +...

MathNote: 1) If y = cosx ; then dy/dx = sinx 2) If y = x^2 ; then dy/dx = 2x 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) Now, the given function is...

MathNote: 1) If y = cosx ; then dy/dx = sinx 2) If y = sinx ; then dy/dx = cosx 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y = k ; where 'k'...

MathNo, sin^2 (x) is the square of sin(x), or sin(x) taken to the second power. By definition, this means sin(x) multiplied by itself two times: `sin^2(x) = sin(x)*sin(x)` . For example, if x = 30...

MathSlope of a tangent line to a function at a point is equal to the derivative of the function at that point. `y=(1+x^3)^(1/2)` `dy/dx=(1/2)*(1+x^3)^((1/2)1)*3x^2` `dy/dx=(3x^2)/(2(sqrt(1+x^3)))`...

Math`y = pi  x` is the exact answer. At `x = pi` `` `y = sin(sin(pi)) = sin(0) = 0` `y' = 1/2 (cos(x  sin(x)) + cos(x + sin(x))) = 1/2(cos(pi) + cos (pi))` =1. So y = x + pi, using the slope...

Mathy = x `y' = cos(x) + 2sin(x)cos(x).` x = 0 `y' = 1 + 0 = 1` Substituting into the yintercept form of the line, `y = 1x + 0 = x`

MathLet `u = sin(pi*x)` `y = 2^(u), (dy)/(du) = 2^(u)log2` `(du)/(dx) = cos(pi*x) * pi` and `y' = pi*log(2)*2^(sin(pi*x))*cos(pi*x)`

Math`y=x^2e^(1/x)` Derivative can be found by using the product rule `y'= x^2*(e^(1/x)(1*1*x^2)) +e^(1/x)*2x` `y'= x^2(x^2e^(1/x) )+ 2xe^(1/x)` `y' = e^(1/x) +2xe^(1/x)` `y'=(1+2x)e^(1/x)`

Math`d/(dy) [cos((1e^(2x))/(1+e^(2x)))] = sin((1e^(2x))/(1+e^(2x)))[d/(dx) ((1e^(2x))/(1+e^(2x)))]` ` ` `= sin((1e^(2x))/(1+e^(2x))){((1+e^(2x))(d/dx)(1e^(2x)) ...

Math`y=sqrt(1+xe^(2x))` `y'=(1/2)*((1+xe^(2x))^((1/2)1)) *d/dx (sqrt(1+xe^(2x)))` `y'=(1/(2sqrt(1+xe^(2x)))) *(xd/dx e^(2x) +e^(2x)d/dx x)` `y'=(1/(2sqrt(1+xe^(2x)))) *(xe^(2x)*(2) +...

Math`y' = d/(d(theta))cot^2(sin(theta))` `= 2 cot (sin(theta)) * d/(d(theta)) [cot(sin(theta))]` `= 2cot(sin(theta))*(csc^2(sin(theta)))*d/(d(theta)) (sin(theta))` `= 2...

Math`y'= d/dx e^(ktan(sqrt(x)))` `= e^(ktan(sqrt(x))) d/dx (ktan(sqrt(x)))` `= k e^(ktan (sqrt(x))) * sec^2(sqrt(x)) * d/dx (sqrt(x))` `= k*e^(ktan(sqrt(x))) * sec^2(sqrt(x))*(1/2 x^(1/2))` `=...

MathUsing d/dt (e^t) = e^t and d/dt (tan t) = sec^2 t `y' = d/dt (tan e^t) + d/dt e^(tan t)` `= sec^2 (e^t) d/dt e^t + e^(tan t) d/dt (tan t)` `= e^t sec^2(e^t) + e^(tan t) sec^2 t` hope this helps.

MathUsing d/dx (sin x) = cos x . d/dx (x) = cos x `y' = d/dx sin(sin(sin x)) = cos(sin(sinx)) * d/dx (sin(sinx))` `= cos(sin(sin x)) * cos(sin x) d/dx (sin x)` `= cos(sin(sin x)) * cos(sin x) * cos...

Math`y' = d/dt [sin^2(e^(sin^2t))]` `= 2 sin(e^(sin^2t))*d/dt(sin(e^(sin^2t)))` `= 2 sin(e^(sin^2t))* cos(e^(sin^2t))*d/dt (e^(sin^2t)))` `= 2 sin(e^(sin^2t))*cos(e^(sin^2t))*e^(sin^2t)*2sint*cost` `=...

Math`y=sqrt(x+sqrt(x+sqrt(x)))` `y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *d/dx (x+sqrt(x+sqrt(x)))` `y'=(1/(2sqrt(x+sqrt(x+sqrt(x))))) *(1+d/dxsqrt(x+sqrt(x)))` `y'=(1/(2sqrt(x+sqrt(x+sqrt(x)))))...

Math`g'(x) = d/dx (2ra^(rx)+n)^p` `= p(2ra^(rx)+n)^(p1) * d/dx (2ra^(rx)+n)` `= p(2ra^(rx)+n)^(p1) * 2r*ln a*a^(rx)*r` `= 2pr^2 a^(rx)*ln(a)*(2ra^(rx)+n)^(p1)` Hope this helps.

Math`y'=d/dx ((2^3)^(x))^2 = ln(2) * ((2^3)^x)^2 d/dx ((3^x)^2)` `= ln(2) ln(3) *(3^x)^2 * d/dx (x^2) * ((2^3)^x)^2` `= ln(2)* ln(3) * 2x * (3^x)^2 * ((2^3)^x)^2 ` `= ln(2)*ln(3)*x*(3^x)^2 *...

MathNotes: 1) If y = cos(ax) ; then dy/dx = a*sin(ax) ; where 'a' = constant 2) If y = tan(ax) ; then dy/dx = a*sec^2(ax) 3) If y = sin(ax) ; then dy/dx = a*cos(ax) ; where 'a' = constant 4) If y =...

Math`y=(x+(x+sin^2x)^3)^4` `y'=4(x+(x+sin^2x)^3)^(41)*((1+3(x+sin^2x)^(31)*(1+2sinxcosx))` `y'=4(x+(x+sin^2x)^3)^3 *((1+3(x+sin^2x)^2 *(1+sin2x))`

Math`dy/dx=(sin(x^2))*(2x)` ` `y'= 2x*sin(x^2) Derivative of the product of a function can be found by d/dx(uv) = u*dv/dx +v*du/dx y" = `2(x*d/dxsin(x^2) + (sin(x^2))*d/dx(x))` `y'' =...

Math`y=cos^2x` `y'=(2cosx)(sinx)` `y'=2sinxcosx` `y" = 2((sinx)(sinx) +cosxcosx)` ` `y"=`2(sin^2x +cos^2x)` y"=`2sin^2x2cos^2x`

MathUsing the product rule: d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x) `y'= d/dx [e^(alpha x) sin (beta x)]` `= e^(alpha x) * alpha * sin(beta x) + e^(alpha x) * cos(beta x)* beta` `= e^(alpha x) *[alpha...

Math`y' = d/dx e^(e^x)` `= e^(e^x) * d/dx(e^x) = e^(e^x)*e^x` `y'' = d/dx(e^(e^x)*e^x)` `= d/dx(e^(e^x)) * e^x + e^(e^x) *d/dx e^x` `= e^(e^x)*e^(2x) + e^x*e^(e^x)` `= e^(e^x)*e^x (e^x +1)` ``hope this...

MathSlope of the tangent line to the curve at the given point is equal to the derivative of the function at that point. `y=(1+2x)^10` `dy/dx=10(1+2x)^(101)*2` `dy/dx=20(1+2x)^9` Therefore the slope...

Math`f(t)=sin(e^t)+e^(sint)` `f'(t)=cos(e^t).e^t+e^(sint).cost`

Math`y=cos(a^3+x^3)` `y'=sin(a^3+x^3).(3x^2)` `y'=3x^2sin(a^3+x^3)`

MathAssuming 'a' is independent of x, since nothing else is given. `y' = d/dx (a^3) + d/dx(cos^3(x))` `= 0 + 3*cos^2(x)*d/dx(cos(x))` `= 3 sin(x) cos^2(x)` hope this helps

Mathy=xe^(kx) y'=e^(kx)(dx/dx)+x(d(e^(kx))/dx) y'=e^(kx)+x(k)e^(kx) y'=e^(kx).(1kx)

Math`y=e^(2t)cos(4t)` `y'=(4sin4t)e^(2t)+cos4t.(2)e^(2t)` `y'=2(2sin4t.e^(2t)+cos4t.e^(2t))` ``

MathUsing the product rule: `d/dx[a(x)b(x)] = a'(x)b(x) + a(x)b'(x)` we get: `f'(x) = 4(2x3)^3*(2)(x^2+x+1)^5 ` `+ (2x3)^4 * 5(x^2+x+1)^4 *(2x+1)` `= (2x3)^3 (x^2+x+1)^4 [8(x^2+x+1) +...

MathUsing the product rule for derivatives: `d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)` `g'(x) = d/dx[(x^2+1)(x^2+2)^6]= 2x*(x^2+2)^6 + (x^2+1)*6(x^2+2)^5*2x` `= (x^2+2)^5 * 2x*[(x^2+2)+6*(x^2+1)] =...

MathUsing the product rule of derivative, we can solve this as: `h'(t) = (2/3) (t+1)^(1/3) *(1) * (2t^21)^3 ` `+ (t+1)^(2/3) * 3*(2t^21)^2 * (4t)` `= (2t^21)^2 (t+1)^(1/3) [ (2/3)(2t^21) +...

MathUsing the quotient rule of derivatives: `d/dx f(x)g(x) = f'(x)g(x) + f(x)g'(x)` `F'(t) = d/dt [(3t1)^4 (2t+1)^(3)] = 4(3t1)^3 * 3* (2t+1)^(3) ` `+ (3t1)^4 * (3) * (2t+1)^(4) * 2` `=(3t1)^3...

Mathusing the quotient rule for derivatives, `d/dx f(x)/g(x) = (f'(x)g(x)g'(x)f(x))/(g(x))^2` `y' = d/dx [(x^2+1)/(x^21)]^3 = [(x^21)^3 *d/dx (x^2+1)^3  (x^2+1)^3 * d/dx (x^21)^3]/((x^21)^3)^2`...

MathNote: 1) If y = x^n ; then dy/dx = n*{x^(n1)} 2) If y = sqrt(x) ; then dy/dx = 1/{2*sqrt(x)} 3) If a function to be differentiated contains subfunctions,then by the rule of differentiation, the...

Math`y' = 1/(2sqrt(1 + 2e^(3x))) * (d/(dx))(1 + 2e^(3x))` `y' = 1/(2sqrt(1 + 2e^(3x))) (3*2e^(3x))` `y' = 3e^(3x)/sqrt(1 +2e^(3x))`

MathUsing the exponential function rule: `[a^(u(x))]' = ln(a)*a^(u(x))*u'(x) ` we get `y' = [10^(1x^2)]'` `= ln(10)*10^(1x^2)*d/dx(1x^2)` `=2x*ln(10)*10^(1x^2)` Hope this helps.

MathUsing the exponential function rule `(a^(u(x)))'= ln(a)*a^(u(x))*u'(x)` `y' = [5^(1/x)]' = ln(5)*5^(1/x)*d/dx(1/x)` `= ln(5)*5^(1/x)*(1/x^2)`

MathNote: 1) If y = x^n ; then dy/dx = n*{x^(n1)} 2) If a function to be differentiated contains subfunctions,then by the rule of differentiation, the last function is differentiated first. 3) If...

Math`y=r/sqrt(r^2+1)` `y'=r*(1/2)(r^2+1)^((1/2)1) (2r) + 1/sqrt(r^2+1)` `y'=r^2(r^2+1)^(3/2) + 1/sqrt(r^2+1)` `y'= r^2/(r^2+1)^(3/2) +1/sqrt(r^2+1)` `y'=...

MathUsing the quotient rule of derivatives: `y' = [d/(du) (e^ue^(u)) * (e^u+e^(u))  (e^ue^(u)) d/(du) (e^u +e^(u))]/(e^u+e^(u))^2` ` ` `= [(e^u e^(u) *(1))*(e^u+e^(u)) ...

MathNote: 1) If y = sint ; then dy/dt = cost 2) If y = e^(at) ; then dy/dt = a*e^(at) ; where a = constant 3) If y = t^n ; then dy/dt = n*t^(n1) ; where n = real number Now, F(t) = e^[t*sin(2t)]...

MathNote: 1) If y = x^n ; then dy/dx = n*{x^(n1)} 2) If a function to be differentiated contains subfunctions,then by the rule of differentiation, the last function is differentiated first. 3) If...

MathNote: 1)If y = sin(ax) ; then dy/dx = a*cos(ax) 2) If y = tan(ax) ; then dy/dx = a*sec^2(ax) Thus, y = sin(tan(2x)) dy/dx = y' = cos(tan(2x))*[sec^2(2x)]*2 or, dy/dx = y' = 2*cos(tan(2x))*[sec^2(2x)]

MathNote: 1) If y = secx ; then dy/dx = sex*tanx 2) If y = x^n ; then dy/dx = n*{x^(n1)} ; where n = real number 3)If y = a*x ; then dy/dx = a ; where a = constant Now, Let theta be represented as...

MathLet y=f(g(x)) y'=f'(g(x)).g'(x) In the given question `y=(1+4x)^(1/3).` Here `f(x)=(x)^(1/3)` and `g(x)=1+4x` `y'=(1/3)(1+4x)^(1/31).(4)` `=4/(3(1+4x)^(2/3))`

MathGiven function `y=(2x^3+5)^4` This is in the form `y=f(g(x))` Here `f(x)=x^4` and `g(x)=2x^3+5` `y'=4(2x^3+5)^3.(6x^2)` `y'=24x^2(2x^3+5)^3` ``