Finding Dy/dx And Tangent Line Equation For 57x^6 + 10x^95y + Y^9 = 68

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This article provides a step-by-step solution for finding the derivative dydx{\frac{dy}{dx}} of the equation 57x6+10x95y+y9=68{57x^6 + 10x^{95}y + y^9 = 68} using implicit differentiation. Additionally, we will determine the equation of the tangent line to the curve at the point (1, 1), expressing the answer in slope-intercept form (y = mx + b).

Implicit Differentiation and Finding dydx{\frac{dy}{dx}}

To find dydx{\frac{dy}{dx}} for the equation 57x6+10x95y+y9=68{57x^6 + 10x^{95}y + y^9 = 68}, we will employ the technique of implicit differentiation. This method is essential when dealing with equations where y is not explicitly defined as a function of x. Here's a detailed breakdown of the process:

  1. Differentiate both sides of the equation with respect to x:

    Applying the differentiation operator ddx{\frac{d}{dx}} to both sides of the equation, we get: ddx(57x6+10x95y+y9)=ddx(68){\frac{d}{dx}(57x^6 + 10x^{95}y + y^9) = \frac{d}{dx}(68)}

  2. Apply the sum/difference rule and constant multiple rule:

    We can differentiate each term separately: ddx(57x6)+ddx(10x95y)+ddx(y9)=ddx(68){\frac{d}{dx}(57x^6) + \frac{d}{dx}(10x^{95}y) + \frac{d}{dx}(y^9) = \frac{d}{dx}(68)} The constant multiple rule allows us to bring constants outside the derivative: 57ddx(x6)+10ddx(x95y)+ddx(y9)=ddx(68){57\frac{d}{dx}(x^6) + 10\frac{d}{dx}(x^{95}y) + \frac{d}{dx}(y^9) = \frac{d}{dx}(68)}

  3. Differentiate each term:

    • For the first term, ddx(x6){\frac{d}{dx}(x^6)}, we use the power rule, which states ddx(xn)=nxn−1{\frac{d}{dx}(x^n) = nx^{n-1}}. So, ddx(x6)=6x5{\frac{d}{dx}(x^6) = 6x^5} Thus, the first term becomes: 57â‹…6x5=342x5{57 \cdot 6x^5 = 342x^5}

    • For the second term, ddx(x95y){\frac{d}{dx}(x^{95}y)}, we need to use the product rule, which states ddx(uv)=u′v+uv′{\frac{d}{dx}(uv) = u'v + uv'}. Here, let u=x95{u = x^{95}} and v=y{v = y}. u′=ddx(x95)=95x94{u' = \frac{d}{dx}(x^{95}) = 95x^{94}} v′=ddx(y)=dydx{v' = \frac{d}{dx}(y) = \frac{dy}{dx}} Applying the product rule, we get: ddx(x95y)=95x94y+x95dydx{\frac{d}{dx}(x^{95}y) = 95x^{94}y + x^{95}\frac{dy}{dx}} Thus, the second term becomes: 10(95x94y+x95dydx)=950x94y+10x95dydx{10(95x^{94}y + x^{95}\frac{dy}{dx}) = 950x^{94}y + 10x^{95}\frac{dy}{dx}}

    • For the third term, ddx(y9){\frac{d}{dx}(y^9)}, we use the chain rule. The chain rule states ddx(f(g(x)))=f′(g(x))g′(x){\frac{d}{dx}(f(g(x))) = f'(g(x))g'(x)}. Here, let f(u)=u9{f(u) = u^9} and u=y{u = y}. f′(u)=9u8{f'(u) = 9u^8} g′(x)=dydx{g'(x) = \frac{dy}{dx}} Applying the chain rule, we get: ddx(y9)=9y8dydx{\frac{d}{dx}(y^9) = 9y^8\frac{dy}{dx}}

    • For the right-hand side, ddx(68){\frac{d}{dx}(68)}, since 68 is a constant, its derivative is 0: ddx(68)=0{\frac{d}{dx}(68) = 0}

  4. Combine the differentiated terms:

    Putting it all together, we have: 342x5+950x94y+10x95dydx+9y8dydx=0{342x^5 + 950x^{94}y + 10x^{95}\frac{dy}{dx} + 9y^8\frac{dy}{dx} = 0}

  5. Isolate dydx{\frac{dy}{dx}} terms:

    Move all terms containing dydx{\frac{dy}{dx}} to one side and the remaining terms to the other side: 10x95dydx+9y8dydx=−342x5−950x94y{10x^{95}\frac{dy}{dx} + 9y^8\frac{dy}{dx} = -342x^5 - 950x^{94}y}

  6. Factor out dydx{\frac{dy}{dx}}:

    Factor out dydx{\frac{dy}{dx}} from the left side: dydx(10x95+9y8)=−342x5−950x94y{\frac{dy}{dx}(10x^{95} + 9y^8) = -342x^5 - 950x^{94}y}

  7. Solve for dydx{\frac{dy}{dx}}:

    Divide both sides by (10x95+9y8){(10x^{95} + 9y^8)} to solve for dydx{\frac{dy}{dx}}: dydx=−342x5−950x94y10x95+9y8{\frac{dy}{dx} = \frac{-342x^5 - 950x^{94}y}{10x^{95} + 9y^8}}

    Thus, we have found the derivative dydx{\frac{dy}{dx}}.

Finding the Equation of the Tangent Line at (1, 1)

Now that we have the expression for dydx{\frac{dy}{dx}}, we can find the equation of the tangent line to the curve at the point (1, 1). The tangent line to a curve at a given point represents the line that best approximates the curve at that point. Here are the steps:

  1. Calculate the slope (m) at (1, 1):

    The slope of the tangent line at a point is given by the value of the derivative dydx{\frac{dy}{dx}} at that point. Substitute x = 1 and y = 1 into the expression for dydx{\frac{dy}{dx}}: dydx∣(1,1)=−342(1)5−950(1)94(1)10(1)95+9(1)8{\frac{dy}{dx}|_{(1,1)} = \frac{-342(1)^5 - 950(1)^{94}(1)}{10(1)^{95} + 9(1)^8}} dydx∣(1,1)=−342−95010+9{\frac{dy}{dx}|_{(1,1)} = \frac{-342 - 950}{10 + 9}} dydx∣(1,1)=−129219{\frac{dy}{dx}|_{(1,1)} = \frac{-1292}{19}} m=−68{m = -68} Thus, the slope of the tangent line at (1, 1) is -68.

  2. Use the point-slope form to find the equation of the tangent line:

    The point-slope form of a line is given by: y−y1=m(x−x1){y - y_1 = m(x - x_1)} where (x1,y1){(x_1, y_1)} is a point on the line and m is the slope. In our case, (x1,y1)=(1,1){(x_1, y_1) = (1, 1)} and m=−68{m = -68}. Substituting these values, we get: y−1=−68(x−1){y - 1 = -68(x - 1)}

  3. Convert to slope-intercept form (y = mx + b):

    To express the equation in slope-intercept form, we need to solve for y: y−1=−68x+68{y - 1 = -68x + 68} y=−68x+68+1{y = -68x + 68 + 1} y=−68x+69{y = -68x + 69}

    Therefore, the equation of the tangent line to the curve at (1, 1) is y=−68x+69{y = -68x + 69}.

Summary

In summary, we have successfully found the derivative dydx{\frac{dy}{dx}} of the equation 57x6+10x95y+y9=68{57x^6 + 10x^{95}y + y^9 = 68} using implicit differentiation. The derivative is given by: dydx=−342x5−950x94y10x95+9y8{\frac{dy}{dx} = \frac{-342x^5 - 950x^{94}y}{10x^{95} + 9y^8}}

We then calculated the equation of the tangent line to the curve at the point (1, 1) by first finding the slope at that point and then using the point-slope form to derive the equation. The equation of the tangent line is: y=−68x+69{y = -68x + 69}

This process demonstrates the application of implicit differentiation and tangent line calculations in calculus. Understanding these concepts is crucial for solving various problems in calculus and related fields.