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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` ``

MathThe given function is `y=tan(pix)` This is in the form `y=f(g(x))` Here `f(x)=tanx` and `g(x)=pix` `dy/dx=sec^2(pix).pi` `=pisec^2(pix)`

MathThe given function is `y=sin(cotx)` This is in the form `y=f(g(x))` `f(x)=sinx and g(x)=cotx` `dy/dx=cos(cotx).(cosec^2x)` `=cos(cotx)cosec^2x`

MathThe given function is `y=e^(sqrt(x))` This is in the form `y=f(g(x))` Here `f(x)=e^x and g(x)=sqrt(x)` `dy/dx=e^(sqrt(x))(1/(2sqrt(x)))` =`(e^(sqrt(x)))/(2sqrt(x))`

MathThe given function is `y=sqrt(2e^x)` This is in the form `y=f(g(x))` Here `f(x)=sqrt(x)` and `g(x)=2e^x` `y'=(1/(2sqrt(2e^x))).(e^x)` `y'=(e^x)/(2sqrt(2e^x))`

Math`F(x)=(x^4+3x^22)^5` `F'(x)=5(x^4+3x^22)^(51) *(4x^(41)+3*2*x^(21))` `F'(x)= 5(x^4+3x^22)^4 *(4x^3+6x)` `F'(x) = 10(2x^3+3x)(x^4+3x^22)^4`

Math`F(x)=(4xx^2)^100` `F'(x)=100(4xx^2)^(1001) *(42x)` `F'(x)=100(4xx^2)^99(42x)` `F'(x) = 200(2x)(4xx^2)^99`

MathF(x) = `sqrt(12x)` `F'(x) = (1/2)*(12x)^((1/2)1)*(2)` `F'(x) = (1)/sqrt(12x)`

MathThe given function is `f(x)=(1+secx)^(2)` `f'(x)=(2)(1+secx)^(21)(secx.tanx)` `f'(x)=(2secx.tanx)/(1+secx)^3` ``

Mathf(z)=`1/(z^2+1)` `f'(z) = 1((z^2+1)^(11))*2z` `f'(z) = 1(z^2+1)^2*2z` `f'(z) = (2z)/(z^2+1)^2`

MathThe slope of a line tangent to the curve y = f(x) at a point where x = a, is given by the value of f'(a). In the problem, the curve is represented by y = 2*e^x + 5x^3 + 3x. First determine the...

Math`lim_(x > 1) sin(x  1)/(x^2 + x  2)` `=lim_((x1) > 0) sin(x  1)/((x+2)(x1))` let `t= x1` so, `lim_((t) > 0) sin(t)/((t+3)(t))` As we know that as `t>0 sin(t) = t` , so...

Math`y = (1  sec(x))/tan(x)` `=(1 /tan(x)) (sec(x)/tan(x))` `y= cot(x)  csc(x)` so now the y' is as follows `y' = d/(dx) (cot(x)  csc(x))` `= d/(dx) (cot(x))  d/(dx)(csc(x))` as `d/(dx)...

MathTo find `y'=d/(dx) x*e^x csc(x)` let `a= x*e^x` so, `y'=d/(dx) (a*csc(x))` `= (d/(dx) a) *csc(x) + a* d/(dx)(csc(x))` substituting` a = x*e^x` so` a' = (d/(dx) a)` `= (d/(dx) x*e^x) = x*e^x + e^x`...

Mathy = x^2 sin(x) tan(x) to find y' so, y' = (x^2 sin(x) tan(x))' let a= sin(x) tan(x) so, y' = (x^2 * a)' = x^2 (a)'+ a *(2x) where a' = ( sin(x) tan(x))' = sin(x) d/(dx) (tan(x))+...

Math`d/(dx) (csc(x))` `=d/(dx) (1/sin(x))` `= ((d/(dx) sin x)*1  sinx * d/dx(1))/(sin^2(x))` `= cosx/(sin^2(x))` `=csc x cot(x)`

Math`d/(dx) (sec(x))` `=d/(dx) (1/cos(x))` `= ((d/(dx) cos x)*1  cosx * d/dx(1))/(cos^2(x))` `= sinx/cos^2(x)` `= sec(x) tan(x)`

Math`d/(dx) (cot(x))` `=d/(dx) (cos(x)/sin(x))` `= ( ((cosx *(d/(dx) sin x)) ((d/(dx) cos x * sinx)))/(sin^2 x)` `= (cos^2 x sin^2 x)/(sin^2 x)` `=  1/(sin^2 x)` `= csc^2(x)`

MathBy using the definition of the derivative, `lim_(h > 0) [ cos(x + h)  cos(x) ] / h` `=lim_(h > 0)[ cos(x)cos(h)  sin(x)sin(h)  cos(x) ] / h` `=lim_(h > 0)[ cos(x) [ cos(h)  1 ] ...

MathThe slope of the tangent is equal to the derivative of the equation of the curve at the given point. `dy/dx = secx.tanx` The slope of the tangent is equal to `2sqrt3.` The equation of the tangent...

MathThe slope of tangent to `y=e^xcosx` at (0,1) = `dy/dx` at (0,1) `dy/dx = e^x(cosxsinx)` Slope is equal to 1. The equation of the tangent is equal to `y1=1(x0).` That is `xy+1=0.` ``

MathThe slope of the tangent = `dy/dx` at (pi,1). `dy/dx = sinxcosx` The slope of the tangent is equal to 1. The equation of the tangent is `y(1)=1(xpi)` . that is `xy=pi+1` ``

MathThe slope of the tangent is equal to the derivative of the equation of the curve at the given point. `dy/dx = 1+sec^2x` Slope of the tangent is equal to 2 The equation of the tangent is equal to...

Math`lim_(x > 0) sin(3x)/x` as when x>0 we get `sin(nx)` thends to be equal to `"nx"` so, we get `lim_(x > 0) sin(3x)/x` `=lim_(x > 0) (3x)/x` = 3

Math`lim_(x > 0) sin(4x)/sin(6x)` as when x>0 we get ` sin(nx)` thends to be equal to ` so, we get `lim_(x > 0) sin(4x)/sin(6x)` `=lim_(x > 0) (4x)/(6x)` `= 4/6 ` `= 2/3`

Math`lim_(t >0) tan(6t)/sin(2t)` as when t>0 we get `tan(nt)= sin(nt)` thends to be equal to `"nt"` so, we get `lim_(t >0) tan(6t)/sin(2t)` `=lim_(t >0) (6t)/(2t)` `= 3`

Math`lim_(theta >0) (cos(theta)  1)/sin(theta)` using L'Hopital's rule we get so, `lim_(theta >0) ((cos(theta)  1)')/(sin(theta)')` `=lim_(theta >0) (sin(theta)/cos(theta))` as...

Math`lim_(x>0) sin(3x)/(5x^3  4x)` using LHopital's rule we get `lim_(x>0) sin(3x)/(5x^3  4x)` `=lim_(x>0) (3*cos(3x))/(15x^2  4)` Now as `x>0` , we get `=(3*cos(3*0))/(15(0)^2 ...

Math`lim_(x>0) (sin(3x) sin(5x))/x^2` As x>0 then `sin(nx) = nx` so, `=lim_(x>0) (sin(3x) sin(5x))/x^2` `=lim_(x>0) ((3x) (5x))/x^2` =lim_(x>0) ((3*5*x^2))/x^2 = 15

Math`lim_(theta > 0) sin(theta)/(theta + tan(theta))` As `theta > 0` then `sin(theta)> theta` and also `tan(theta)` so, `lim_(theta > 0) sin(theta)/(theta + tan(theta))`...

Math`lim_(x > 0) sin(x^2)/x` As `x>0` the `sin(x) > x` so, ` =>lim_(x > 0) sin(x^2)/x` `=> lim_(x > 0) x^2/x` `=>lim_(x > 0) x = 0`

Math`lim_(x > pi/4) (1  tan(x))/(sin(x)  cos(x))` Using the L'Hopital's Rule we get `=lim_(x > pi/4) ((1  tan(x))')/((sin(x)  cos(x))')` As `d/dx (1  tan(x)) = sec^2(x)` and `d/dx...

Math`f'(x)=3(2x)2(sinx)` `f'(x)=6x+2sinx`

Math`f'(x)=sinx(1/(2sqrtx))+sqrtx(cosx)` `f'(x)=(sinx+2xcosx)/(2sqrtx)`

Math`f'(x)=((dsinx)/dx)+(1/2)((dcotx)/dx)` `f'(x)=cosx(cosec^2(x))/2`

Math`y'=2(secx.tanx)(cscx.cotx)` `y'=2secx.tanx+cscx.cotx` ``

Math`y'=(sectheta.tantheta)tantheta+sectheta(sec^2theta)` `y'=sectheta(tan^2theta+sec^2theta)`

Math`g(theta)=e^(theta)tanthetathetae^(theta)` `g'(theta)=e^(theta)tantheta+e^(theta)sec^2theta(thetae^(theta)+e^(theta))` `g'(theta)=e^(theta)(tantheta+sec^2thetatheta1)`

Math`y'=c(sint)+2tsint+t^2cost` `y'=(2tc)sint+t^2cost`

Math`f(t)=cot(t)e^(t)` `f'(t)=(cosec^2(t))e^(t)+cot(t)(e^(t))` `f'(t)=((cosec^2(t)+cot(t)))/e^t`

Math`y'=(1*(2tanx)x(sec^2(x)))/(2tanx)^2` `y'=(xsec^2xtanx+2)/(2tanx)^2`

Math`y=(2sintheta.costheta)/2` `y=sin(2theta)/2` `y'=2cos(2theta)/2=cos(2theta)`

Math`f(theta) = sec(theta)/(1 + sec(theta))` so, `f'(theta) = (sec(theta)/(1 + sec(theta)))'` `= (1 (1/(1 + sec(theta))))'` `= 0 (1/(1 + sec(theta)))'` `=  tan(theta)* sec(theta)/(1 +...

MathQuotient Rule: `y'=(((sin(x))*(1sin(x))(cos(x))(cos(x)))/((1sin(x))^(2))) ` `y'=(((cos^(2)(x))+(sin(x))*(sin(x)1))/((sin(x)1)^(2)))` This can simplify to `y' = 1/(1  sin(x))`

Math`y = (t sin(t))/(1 + t)` `y' = ((t sin(t))/(1 + t))'` `= ((1+t) *(d/(dt) (t sin t))  t sint *(d/(dt) ( 1+t)))/(1+t)^2` `=(((1+t) *(t(cos t)+ sint))  (t sint *(1)))/(1+t)^2` ` = (sin(t)...

MathThis is an equation of the area of a circle with a radius of 'r'. 3.14 represents the value of 'pi'. Conventionally, area of a circle is given as: `A = pir^2 = pi/4 d^2` Where A is area, r is...

Math`f'(x)=(5/2)x^(3/2)e^x+x^(5/2)e^x` `f'(x)=(e^x/2)(5x^(3/2)+2x^(5/2))` `f''(x)=(e^x/2)(5x^(3/2)+2x^(5/2))+(e^x/2)((15/2)x^(1/2)+5x^(3/2))` `f''(x)=(e^x/4)(20x^(3/2)+4x^(5/2)+15x^(1/2))`