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Differentiate y = e^x _e x using product rule. Find (dy)/(dx), if x2 + y2 + xy3 = 3 at the given point (1, 1). If y = (t2)/(1+t), x = (t)/(1+t), then find (dy)/(dx).

Mathematics

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Here are the solutions to the differentiation problems:

Part i· 3 steps

Differentiate y=(x2+2)5y = (x^2 + 2)^5

  1. Apply the chain rule.

    Let u=x2+2u = x^2 + 2. Then y=u5y = u^5.

    dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

  2. Differentiate yy with respect to uu and uu with respect to xx.

    dydu=5u4\frac{dy}{du} = 5u^4 dudx=2x\frac{du}{dx} = 2x

  3. Substitute back u=x2+2u = x^2 + 2 and multiply the derivatives.

    dydx=5(x2+2)4⋅(2x)\frac{dy}{dx} = 5(x^2 + 2)^4 \cdot (2x) dydx=10x(x2+2)4\frac{dy}{dx} = \boxed{10x(x^2 + 2)^4}

Part ii· 12 steps

Differentiate y=sin⁡xcos⁡xy = \frac{\sin x}{\cos x}

  1. Recognize that y=sin⁡xcos⁡x=tan⁡xy = \frac{\sin x}{\cos x} = \tan x.

  2. Differentiate y=tan⁡xy = \tan x with respect to xx.

    dydx=sec⁡2x\frac{dy}{dx} = \boxed{\sec^2 x}

    (3a). Differentiate y=exlog⁡exy = e^x \log_e x using product rule.

  3. Identify uu and vv for the product rule, where y=uvy = uv.

    Let u=exu = e^x and v=log⁡ex=ln⁡xv = \log_e x = \ln x.

  4. Find the derivatives of uu and vv with respect to xx.

    dudx=ex\frac{du}{dx} = e^x dvdx=1x\frac{dv}{dx} = \frac{1}{x}

  5. Apply the product rule formula

    dydx=udvdx+vdudx\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.

    dydx=ex(1x)+(ln⁡x)ex\frac{dy}{dx} = e^x \left(\frac{1}{x}\right) + (\ln x) e^x dydx=ex(1x+ln⁡x)\frac{dy}{dx} = \boxed{e^x \left(\frac{1}{x} + \ln x\right)}

    (3b). Find dydx\frac{dy}{dx}, if x2+y2+xy3=3x^2 + y^2 + xy^3 = 3 at the given point (1,1)(1, 1).

  6. Differentiate both sides of the equation with respect to xx using implicit differentiation.

    Remember to apply the product rule for xy3xy^3.

    ddx(x2)+ddx(y2)+ddx(xy3)=ddx(3)\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(xy^3) = \frac{d}{dx}(3) 2x+2ydydx+(1⋅y3+x⋅3y2dydx)=02x + 2y \frac{dy}{dx} + \left(1 \cdot y^3 + x \cdot 3y^2 \frac{dy}{dx}\right) = 0

  7. Rearrange the equation to isolate dydx\frac{dy}{dx}.

    2x+2ydydx+y3+3xy2dydx=02x + 2y \frac{dy}{dx} + y^3 + 3xy^2 \frac{dy}{dx} = 0 (2y+3xy2)dydx=−2x−y3(2y + 3xy^2) \frac{dy}{dx} = -2x - y^3 dydx=−2x−y32y+3xy2\frac{dy}{dx} = \frac{-2x - y^3}{2y + 3xy^2}

  8. Substitute the given point (x,y)=(1,1)(x, y) = (1, 1) into the expression for dydx\frac{dy}{dx}.

    dydx=−2(1)−(1)32(1)+3(1)(1)2\frac{dy}{dx} = \frac{-2(1) - (1)^3}{2(1) + 3(1)(1)^2} dydx=−2−12+3=−35\frac{dy}{dx} = \frac{-2 - 1}{2 + 3} = \frac{-3}{5} dydx=−35\frac{dy}{dx} = \boxed{-\frac{3}{5}}

    (3c). if y=t21+ty = \frac{t^2}{1+t}, x=t1+tx = \frac{t}{1+t}, then find dydx\frac{dy}{dx}

  9. Find dydt\frac{dy}{dt} using the quotient rule.

    Let u=t2u = t^2 and v=1+tv = 1+t.

    dydt=(1+t)(2t)−(t2)(1)(1+t)2=2t+2t2−t2(1+t)2=t2+2t(1+t)2\frac{dy}{dt} = \frac{(1+t)(2t) - (t^2)(1)}{(1+t)^2} = \frac{2t + 2t^2 - t^2}{(1+t)^2} = \frac{t^2 + 2t}{(1+t)^2}

  10. Find dxdt\frac{dx}{dt} using the quotient rule.

    Let u=tu = t and v=1+tv = 1+t.

    dxdt=(1+t)(1)−(t)(1)(1+t)2=1+t−t(1+t)2=1(1+t)2\frac{dx}{dt} = \frac{(1+t)(1) - (t)(1)}{(1+t)^2} = \frac{1+t - t}{(1+t)^2} = \frac{1}{(1+t)^2}

  11. Apply the chain rule for parametric equations

    dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

    dydx=t2+2t(1+t)21(1+t)2\frac{dy}{dx} = \frac{\frac{t^2 + 2t}{(1+t)^2}}{\frac{1}{(1+t)^2}}

  12. Simplify the expression.

    dydx=(t2+2t)⋅(1+t)2(1+t)2=t2+2t\frac{dy}{dx} = (t^2 + 2t) \cdot \frac{(1+t)^2}{(1+t)^2} = t^2 + 2t dydx=t2+2t\frac{dy}{dx} = \boxed{t^2 + 2t}
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