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MathGiven: `x^2+y^2=9` `2x+2ydy/dx=0` `2ydy/dx=2x` `dy/dx=(2x)/(2y)` `dy/dx=(x)/(y)` ``

MathGiven: `x^2y^2=25` `2x2ydy/dx=0` `2ydy/dx=2x` `dy/dx=(2x)/(2y)` `dy/dx=(x)/(y)`

Math`x^(1/2) + y^(1/2) = 16` `or, (1/2)x^(1/2) + (1/2)y^(1/2)(dy/dx) = 0` `or, dy/dx = (x/y)^(1/2) = (y/x)^(1/2)` ``

Math`2x^3 + 3y^3 = 64` ` ` Differentiating both sides w.r.t 'x' we get `2*3*x^2 + 3*3*y^2*(dy/dx) = 0` `or, 2x^2 + 3y^2(dy/dx) = 0` `or, dy/dx = (2/3)*(x/y)^2` ``

MathGiven: `x^3xy+y^2=7` `3x^2[xdy/dx+y(1)]+2ydy/dx=0` `3x^2xdy/dxy+2ydy/dx=0` `xdy/dx+2ydy/dx=3x^2+y` `xdy/dx2ydy/dx=3x^2y` `dy/dx(x+2y)=3x^2y` `dy/dx=(3x^2y)/(x+2y)` `` ``

MathYou need first to evaluate the first derivative of the function, using quotient and chain rules, such that: `f'(x) = 1*((x+4)')/(2sqrt(x+4))` `f'(x) = 1/(2sqrt(x+4))` Now you may evaluate the...

MathWe will need to use chain rule `(f(g(x)))'=f'(g(x))cdot g'(x)` First we find the first derivative. `f'(x)=sin(x^2)cdot 2x=2x sin(x^2)` Now we calculate second derivative, but for that we will...

MathYou need to evaluate second derivative of the given function, at the specified point, hence, you need to evaluate the first derivative, such that: g'(t)=(tan 2t)' => g'(t)=((2t)')/(cos^2 (2t))...

MathYou need to evaluate the derivative of the function, using the chain and quotient rules, such that: `y' = (2(x^2  3x)(x^2  3x)')/((x^2  3x)^4) => y' = (2(x^2  3x)(2x  3))/((x^2  3x)^4)`...

MathFirst we differentiate the function using the following rule `(f/g)'=(f'gfg')/g^2` `f'(t)=(3(t1)(3t+2)\cdot1)/(t1)^2=(3t33t2)/(t1)^2=5/(t1)^2` Now we evaluate the derivative at 0....

MathYou need to evaluate the equation of the tangent line at (9,1), using the formula: f(x)  f(9) = f'(9)(x  1) Notice that f(9) = 1. You need to evaluate f'(x), using the quotient rule, such that:...

MathYou need to evaluate the derivative of the function, using the chain rule, such that: `y' = (26  (sec(4x))^3)' => y' = 3(sec(4x))^2* (sec(4x))*(tan(4x))*(4x)'` `y' = 12(sec(4x))^3*(tan(4x))`...

MathHello! `y(x) = (1/x) + sqrt(cos(x))` . cos(x) must be >=0 and cos(x) < 0 for `pigtxgtpi/2` ` `(at the right neighborhood of `pi/2` ). So there can be only the left derivative. Check that...

Math`f(x)=sqrt(2x^27)` Slope of the tangent line is the derivative of the function at that point. `f'(x)=(1/2)(2x^27)^(1/2)(4x)` `f'(x)=(2x)/sqrt(2x^27)` Slope of the tangent line (m) at (4,5)...

MathThe function `f(x) = (1/3)*x*sqrt(x^2 + 5)` . The slope of a line tangent to this curve at x = a is f'(a). The derivative `f'(x) = (1/3)*(x*sqrt(x^2 + 5))'` = `(1/3)*(x*(sqrt(x^2+5))' +...

MathYou need to find the equation of the tangent line to the given curve, at the point (1,1), using the formula: `f(x)  f(1) = f'(1)(  (1))` You need to put y = f(x) and you need to notice that...

MathThe function `f(x) = (9  x^2)^(2/3)` . At the point (1, 4) the equation of the tangent to this curve is: `(y  4)/(x  1) = f'(1)` `f(x) = (9  x^2)^(2/3)` `f'(x) = (2/3)*(9 ...

MathThe function f(x) = sin(2x). The slope of a line tangent to this curve at the point (pi,0) is given by the equation `(y  0)/(x  pi) = f'(pi)` `f'(x)= (sin (2x))'` = 2*cos(2x) `f'(pi) = 2`...

MathThe function y = cos (3x). The slope of the tangent to the graph of this function at `(pi/4, sqrt2/2)` is `(y + sqrt 2/2)/(x  pi/4) = y'(pi/4)` `y' = 3*sin(3*x)` `y'(pi/4) = 3/sqrt 2` The...

MathYou need to find the equation of the tangent line to the given curve, at the point `(pi/4,1)` , using the formula: `f(x)  f(pi/4) = f'(pi/4)(x  pi/4)` You need to notice that `f(pi/4) = 1.` You...

Math`y = 2(tan(x))^3, ((pi/4),2)` Find an equation of the tangent line to the graph of f at the given...You need to find the equation of the tangent line to the given curve, at the point (pi/4,2), using the formula: `f(x)  f(pi/4) = f'(pi/4)(x  pi/4)` You need to notice that `f(pi/4) = 2.` You need...

Math`f(x) = 5(27x)^4` `f'(x) = 5*4*(7)*(27x)^3` `or, f'(x) = 140(27x)^3` `Thus, f''(x) = 140*3*(7)*(27x)^2` `or, f''(x) = 294*(27x)^2` ``

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, `f(x) = 6(x^3 + 4)^3` `thus, f'(x) = 6*3*3(x^2)*(x^3+4)^2` `or, f'(x) = 54(x^2)(x^3 + 4)^2` Differentiating again w.r.t 'x' we...

MathNote: if y = (ax+b)^n ; where a,b,n are constants ; then dy/dx = an*(ax+b)^(n1) Now, `f(x) = 1/(x6) = (x6)^1` `or, f'(x) = 1(x6)^2` `or, f''(x) = 2(x6)^3` `or, f''(x) = 2/(x6)^3` ` `

MathNote: If y = (ax+b)^n ; where a,b,n are constants ; then dy/dx = na*x^(n1) Now, `y = 8/(x2)^2` `or, y = 8*(x2)^2` `or, y = 16(x2)^3` `or, y = 48(x2)^4` `or, y = 48/(x2)^4` ``

MathNote: 1) If y = sinx ; then dy/dx = cosx 2) If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, `f(x) = sin(x^2)` `f'(x) = 2x*cos(x^2)` `f''(x) = 2*cos(x^2)  4(x^2)*sin(x^2)` ``

Math`f(x) = (sec(pix))^2 = sec^2(pix)` `f'(x) = 2sec(pix)*tan(pix)*sec(pix) = 2pisec^2(pix)*tan(pix)` `f''(x) = 2pi^2*sec^4(pix) + 4pi^2sec^2(pix)*tan^2(pix)` `or, f''(x) = 2pi^2*sec^2(pix)[sec^2(pix)...

MathYou need first to evaluate the first derivative of the function, using chain rule, such that: `h'(x) = ((1/9)(3x + 1)^3)'` `h'(x) = ((3/9)(3x + 1)^2)*(3x + 1)'` `h'(x) = ((1/3)(3x + 1)^2)*(3)`...

MathNote: 1) 2sin(ax)*cos(ax) = sin2(ax) 2) If y = sinx ; then dy/dx = cosx Now, `h(x) = sin(2x)*cos(2x)` `or, h(x) = (1/2)*sin(4x)` `thus, h'(x) = (4/2)cos(4x)` `or, h'(x) = 2cos(4x)` ``

MathGiven the function `g(theta)=sec((theta)/2)tan((theta)/2)` . We have to find its derivative. Here we will apply the product rule i.e., `d/(d theta)(uv)=uv'+u'v` So here we have, `d/(d...

Math`f(x) = cot(x)/sin(x)` `f'(x) = [sin(x)*{cosec^2(x)}  cot(x)*cos(x)]/{sin^2(x)}` `or, f'(x) = [cosec(x)  cot(x)*cos(x)]/{sin^2(x)}` ``

MathGiven: `g(v)=cos(v)/csc(v)` The original function equivalent to `g(v)=cos(v)sin(v).` `g'(x)=cos(v)cos(v)+sin(v)(sin(v))` `g'(x)=cos^2(v)sin^2(v)` ``

Math`y = 4sec^2(x)` `dy/dx = 4*2*sec(x)*sec(x)*tan(x) ` `or, dy/dx = 8sec^2(x)tan(x)` ` `Note: 1) If y = sec(x) ; then dy/dx = secx*tanx

Mathg(t) = `5(cos(pit))^2` `g'(t) = 5*2*cos(pit)*{sin(pit)}*pi` `or, g'(t) = 10cos(pit)*sin(pit)` `or, g'(t) = 5sin(2pit)` `Note:` `1)2sint*cost = sin2t` ``

Math`f(theta) = {tan5(theta)}^2 = tan^2(5theta)` `thus, f'(theta) = 2*tan(5theta)*sec^2(5theta)*5` `or, f'(theta) = 10*tan(5theta)*sec^2(5theta)` ``

Math`g(theta) = {cos(8theta)}^2 = cos^2(8theta)` `g'(theta) = 2*cos(8theta)*(sin(8theta))*8` `or, g'(theta) = 16sin(8theta)cos(8theta)` `or, g'(theta) = 8*sin(16theta)` ``

MathYou need to evaluate the derivative of the function, hence you need to use the chain rule, differentiating first the power, then the sine function and then the argument, such that: f'(theta) =...

Math`h(t) = 2(cot(pit + 2))^2 = 2cot^2(pit + 2)` `h'(t) = 4cot(pit+2)*[cosec^2(pit+2)]*pi` `or, h'(t) = 4pi*cot(pit+2)*cosec^2(pit+2)` ``

Math`f(t) = 3(sec(pit1))^2` `or, f(t) = 3sec^2(pit1)` `thus, f'(t) = 6sec(pit1)*[sec(pit1)*tan(pit1)]*pi` `or, f'(t) = 6pi*sec^2(pit1)*tan(pit1)` ``

Math`y = 3x  5cos(pix)^2` `dy/dx = y' = 3 + 5*2*(pix)*pi*sin(pix)^2` `or, dy/dx = y' = 3 + 10(pi^2)*x*sin(pix)^2` ``

MathYou need to evaluate the derivative of the function, hence you need to use the chain rule, where it is required, such that: `y' = (sqrt x)' + (1/4)*(sin (2x)^2)'*((2x)^2)'*(2x)'` `y' = 1/(2(sqrt...

MathHello! We have the composite functions, for such a functions `[f(g(x))]' = f'(g(x))*g'(x).` Also recall that `[sin(y)]' = cos(y)` and `[y^(1/3)]' = (1/3)*y^(2/3).` Let's start: `y(x) =...

Math`y = sin(tan(2x))` `dy/dx = [(dsin(tan(2x)))/dx]*[(dtan(2x))/dx]*[(d2x)/dx]` `or, dy/dx = y' = [cos(tan(2x))]*[sec^2(2x)]*2` `or, dy/dx = y' = 2sec^2(2x)*cos(tan(2x))` `` ` `

MathGiven the function `y=cos(sqrt(sin(tan(pi x))))` . We have to find the derivative. Let us begin, `(dy)/(dx)=sin(sqrt(sin(tan(pi x)))).d/(dx)[sqrt(sin(tan(pi x)))] >(1)` Now,...

MathYou need to evaluate the derivative of the function, using the chain rule, such that: `y' = ((x^2 + 8x)^(1/2))' => y' = ((1/2)(x^2 + 8x)^(1/2  1))(x^2 + 8x)'` `y' = ((1/2)(x^2 + 8x)^(1/2))(2x...

MathYou need to evaluate the derivative of the function, using the chain rule, such that: `y' = ((3x^3 + 4x)^(1/5))' => y' = ((1/5)(3x^3 + 4x)^(1/5  1))(3x^3 + 4x)'` `y' = ((1/5)(3x^3 +...

MathThe function f(x) = 5/(x^3  2) = 5*(x^3  2)^1. The derivative f'(x) = 5*(x^3  2)^2*1*3x^2 = (15x^2)/(x^3  2)^2 At x = 2, f'(x) = (15*4)/(8  2)^2 = 60/100 = 0.6 The equation of the...

MathGiven: `y=(3/(t2)^4)` Rewrite the equation as `y=3(t2)^4` Find the derivative using the Chain Rule. `y'=12(t2)^5` `y'=12/(t2)^5`

Math`y = 1/sqrt(3x+5) = (3x+5)^(1/2)` `thus, dy/dx = y' = (1/2)*{(3x+5)^(3/2)}*3` `or, dy/dx = (3/2)*(3x+5)^(3/2)` `or, dy/dx = y' = (3/2)*{1/(3x+5)^(3/2)}` ``

Math`g(t) = 1/sqrt(t^2  2)` ` ` `or, g(t) = (t^2  2)^(1/2)` `Thus, g'(t) = (1/2)*(2t)*(t^2  2)^(3/2)` `or, g'(t) = t*(t^2  2)^(3/2)` `or, g'(t) = t/(t^2  2)^(3/2)` `` ` `