Hard A-Level Implicit differentiation Questions

Challenging, exam-style A-Level Implicit differentiation questions with worked solutions. Stretch yourself on the hardest implicit-differentiation, exponential problems.

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A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Which of the following is dydx\dfrac{dy}{dx} for the curve xy=12x y = 12?
Show worked solution

Worked solution

  1. Write down the relation

    xy=12x y = 12

    This implicit equation connects x and y without y being isolated.

  2. Differentiate both sides with respect to xx

    dydxx+y=0\frac{dy}{dx} x + y = 0

    Differentiate term by term, treating y as a function of x.

  3. Collect the terms containing dydx\frac{dy}{dx}

    (x)dydx=−y\left(x\right)\frac{dy}{dx} = - y

    Group every term that has a factor of dy/dx on one side.

  4. Make dydx\frac{dy}{dx} the subject

    dydx=−yx\frac{dy}{dx} = \frac{- y}{x}

    Divide both sides by the coefficient of dy/dx.

  5. Check the point (2, 6)\left(2,\ 6\right) lies on the curve

    12=1212 = 12

    A gradient only makes sense at a point on the curve.

  6. Substitute the point into the numerator of dydx\frac{dy}{dx}

    −y=−6- y = -6

    Evaluate the top of the gradient fraction at the point.

  7. Substitute the point into the denominator of dydx\frac{dy}{dx}

    x=2x = 2

    Evaluate the bottom of the gradient fraction at the point.

  8. Evaluate the gradient at the point

    dydx∣(2, 6)=−3\left.\frac{dy}{dx}\right|_{\left(2,\ 6\right)} = -3

    Divide the numerator value by the denominator value.

  9. Treat yy as a function of xx

    y=y(x)y=y(x)

    In an implicit relation y is not isolated, but it still depends on x.

  10. Apply the chain rule to the yy-terms

    ddxf(y)=f′(y)dydx\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}

    Every function of y picks up a factor of dy/dx when differentiated in x.

  11. Use the product rule on mixed xyxy terms

    ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

    A term containing both x and y needs the product rule.

  12. The derivative of a constant is zero

    ddx(c)=0\frac{d}{dx}(c)=0

    Constant terms disappear when differentiated.

  13. Recall the power rule for xx-terms

    ddxxn=nxn−1\frac{d}{dx}x^{n}=nx^{n-1}

    Pure powers of x differentiate in the usual way.

  14. Differentiate the term xyx y

    ddx(xy)=dydxx+y\frac{d}{dx}\left(x y\right) = \frac{dy}{dx} x + y

    Differentiate this term, using the chain rule if it involves y.

  15. Select the correct derivative

    dydx=−yx\frac{dy}{dx} = - \frac{y}{x}

    This matches the correct option.

Answer
−yx- \frac{y}{x}
Question 2
8 markschallenging
Which of the following is dydx\dfrac{dy}{dx} for the curve x2−xy+y2=7x^{2} - x y + y^{2} = 7?
Show worked solution

Worked solution

  1. Write down the relation

    x2−xy+y2=7x^{2} - x y + y^{2} = 7

    This implicit equation connects x and y without y being isolated.

  2. Differentiate both sides with respect to xx

    −dydxx+2dydxy+2x−y=0- \frac{dy}{dx} x + 2 \frac{dy}{dx} y + 2 x - y = 0

    Differentiate term by term, treating y as a function of x.

  3. Collect the terms containing dydx\frac{dy}{dx}

    (−x+2y)dydx=−2x+y\left(- x + 2 y\right)\frac{dy}{dx} = - 2 x + y

    Group every term that has a factor of dy/dx on one side.

  4. Make dydx\frac{dy}{dx} the subject

    dydx=−2x+y−x+2y\frac{dy}{dx} = \frac{- 2 x + y}{- x + 2 y}

    Divide both sides by the coefficient of dy/dx.

  5. Check the point (1, −2)\left(1,\ -2\right) lies on the curve

    7=77 = 7

    A gradient only makes sense at a point on the curve.

  6. Substitute the point into the numerator of dydx\frac{dy}{dx}

    −2x+y=−4- 2 x + y = -4

    Evaluate the top of the gradient fraction at the point.

  7. Substitute the point into the denominator of dydx\frac{dy}{dx}

    −x+2y=−5- x + 2 y = -5

    Evaluate the bottom of the gradient fraction at the point.

  8. Evaluate the gradient at the point

    dydx∣(1, −2)=45\left.\frac{dy}{dx}\right|_{\left(1,\ -2\right)} = \frac{4}{5}

    Divide the numerator value by the denominator value.

  9. Treat yy as a function of xx

    y=y(x)y=y(x)

    In an implicit relation y is not isolated, but it still depends on x.

  10. Apply the chain rule to the yy-terms

    ddxf(y)=f′(y)dydx\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}

    Every function of y picks up a factor of dy/dx when differentiated in x.

  11. Use the product rule on mixed xyxy terms

    ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

    A term containing both x and y needs the product rule.

  12. The derivative of a constant is zero

    ddx(c)=0\frac{d}{dx}(c)=0

    Constant terms disappear when differentiated.

  13. Recall the power rule for xx-terms

    ddxxn=nxn−1\frac{d}{dx}x^{n}=nx^{n-1}

    Pure powers of x differentiate in the usual way.

  14. Differentiate the term x2x^{2}

    ddx(x2)=2x\frac{d}{dx}\left(x^{2}\right) = 2 x

    Differentiate this term, using the chain rule if it involves y.

  15. Select the correct derivative

    dydx=2x−yx−2y\frac{dy}{dx} = \frac{2 x - y}{x - 2 y}

    This matches the correct option.

Answer
2x−yx−2y\frac{2 x - y}{x - 2 y}
Question 3
8 markschallenging
Which of the following is dydx\dfrac{dy}{dx} for the curve x2+y2=25x^{2} + y^{2} = 25?
Show worked solution

Worked solution

  1. Write down the relation

    x2+y2=25x^{2} + y^{2} = 25

    This implicit equation connects x and y without y being isolated.

  2. Differentiate both sides with respect to xx

    2dydxy+2x=02 \frac{dy}{dx} y + 2 x = 0

    Differentiate term by term, treating y as a function of x.

  3. Collect the terms containing dydx\frac{dy}{dx}

    (2y)dydx=−2x\left(2 y\right)\frac{dy}{dx} = - 2 x

    Group every term that has a factor of dy/dx on one side.

  4. Make dydx\frac{dy}{dx} the subject

    dydx=−2x2y\frac{dy}{dx} = \frac{- 2 x}{2 y}

    Divide both sides by the coefficient of dy/dx.

  5. Check the point (3, 4)\left(3,\ 4\right) lies on the curve

    25=2525 = 25

    A gradient only makes sense at a point on the curve.

  6. Substitute the point into the numerator of dydx\frac{dy}{dx}

    −2x=−6- 2 x = -6

    Evaluate the top of the gradient fraction at the point.

  7. Substitute the point into the denominator of dydx\frac{dy}{dx}

    2y=82 y = 8

    Evaluate the bottom of the gradient fraction at the point.

  8. Evaluate the gradient at the point

    dydx∣(3, 4)=−34\left.\frac{dy}{dx}\right|_{\left(3,\ 4\right)} = - \frac{3}{4}

    Divide the numerator value by the denominator value.

  9. Treat yy as a function of xx

    y=y(x)y=y(x)

    In an implicit relation y is not isolated, but it still depends on x.

  10. Apply the chain rule to the yy-terms

    ddxf(y)=f′(y)dydx\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}

    Every function of y picks up a factor of dy/dx when differentiated in x.

  11. Use the product rule on mixed xyxy terms

    ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

    A term containing both x and y needs the product rule.

  12. The derivative of a constant is zero

    ddx(c)=0\frac{d}{dx}(c)=0

    Constant terms disappear when differentiated.

  13. Recall the power rule for xx-terms

    ddxxn=nxn−1\frac{d}{dx}x^{n}=nx^{n-1}

    Pure powers of x differentiate in the usual way.

  14. Differentiate the term x2x^{2}

    ddx(x2)=2x\frac{d}{dx}\left(x^{2}\right) = 2 x

    Differentiate this term, using the chain rule if it involves y.

  15. Select the correct derivative

    dydx=−xy\frac{dy}{dx} = - \frac{x}{y}

    This matches the correct option.

Answer
−xy- \frac{x}{y}
Question 4
8 markschallenging
Which of the following is dydx\dfrac{dy}{dx} for the curve exy=e6e^{x y} = e^{6}?
Show worked solution

Worked solution

  1. Write down the relation

    exy=e6e^{x y} = e^{6}

    This implicit equation connects x and y without y being isolated.

  2. Differentiate both sides with respect to xx

    dydxxexy+yexy=0\frac{dy}{dx} x e^{x y} + y e^{x y} = 0

    Differentiate term by term, treating y as a function of x.

  3. Collect the terms containing dydx\frac{dy}{dx}

    (xexy)dydx=−yexy\left(x e^{x y}\right)\frac{dy}{dx} = - y e^{x y}

    Group every term that has a factor of dy/dx on one side.

  4. Make dydx\frac{dy}{dx} the subject

    dydx=−yexyxexy\frac{dy}{dx} = \frac{- y e^{x y}}{x e^{x y}}

    Divide both sides by the coefficient of dy/dx.

  5. Check the point (2, 3)\left(2,\ 3\right) lies on the curve

    e6=e6e^{6} = e^{6}

    A gradient only makes sense at a point on the curve.

  6. Substitute the point into the numerator of dydx\frac{dy}{dx}

    −yexy=−3e6- y e^{x y} = - 3 e^{6}

    Evaluate the top of the gradient fraction at the point.

  7. Substitute the point into the denominator of dydx\frac{dy}{dx}

    xexy=2e6x e^{x y} = 2 e^{6}

    Evaluate the bottom of the gradient fraction at the point.

  8. Evaluate the gradient at the point

    dydx∣(2, 3)=−32\left.\frac{dy}{dx}\right|_{\left(2,\ 3\right)} = - \frac{3}{2}

    Divide the numerator value by the denominator value.

  9. Treat yy as a function of xx

    y=y(x)y=y(x)

    In an implicit relation y is not isolated, but it still depends on x.

  10. Apply the chain rule to the yy-terms

    ddxf(y)=f′(y)dydx\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}

    Every function of y picks up a factor of dy/dx when differentiated in x.

  11. Use the product rule on mixed xyxy terms

    ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

    A term containing both x and y needs the product rule.

  12. The derivative of a constant is zero

    ddx(c)=0\frac{d}{dx}(c)=0

    Constant terms disappear when differentiated.

  13. Recall the power rule for xx-terms

    ddxxn=nxn−1\frac{d}{dx}x^{n}=nx^{n-1}

    Pure powers of x differentiate in the usual way.

  14. Differentiate the term exye^{x y}

    ddx(exy)=dydxxexy+yexy\frac{d}{dx}\left(e^{x y}\right) = \frac{dy}{dx} x e^{x y} + y e^{x y}

    Differentiate this term, using the chain rule if it involves y.

  15. Select the correct derivative

    dydx=−yx\frac{dy}{dx} = - \frac{y}{x}

    This matches the correct option.

Answer
−yx- \frac{y}{x}
Question 5
8 markschallenging
Which of the following is dydx\dfrac{dy}{dx} for the curve x2+xy+y2=7x^{2} + x y + y^{2} = 7?
Show worked solution

Worked solution

  1. Write down the relation

    x2+xy+y2=7x^{2} + x y + y^{2} = 7

    This implicit equation connects x and y without y being isolated.

  2. Differentiate both sides with respect to xx

    dydxx+2dydxy+2x+y=0\frac{dy}{dx} x + 2 \frac{dy}{dx} y + 2 x + y = 0

    Differentiate term by term, treating y as a function of x.

  3. Collect the terms containing dydx\frac{dy}{dx}

    (x+2y)dydx=−2x−y\left(x + 2 y\right)\frac{dy}{dx} = - 2 x - y

    Group every term that has a factor of dy/dx on one side.

  4. Make dydx\frac{dy}{dx} the subject

    dydx=−2x−yx+2y\frac{dy}{dx} = \frac{- 2 x - y}{x + 2 y}

    Divide both sides by the coefficient of dy/dx.

  5. Check the point (1, 2)\left(1,\ 2\right) lies on the curve

    7=77 = 7

    A gradient only makes sense at a point on the curve.

  6. Substitute the point into the numerator of dydx\frac{dy}{dx}

    −2x−y=−4- 2 x - y = -4

    Evaluate the top of the gradient fraction at the point.

  7. Substitute the point into the denominator of dydx\frac{dy}{dx}

    x+2y=5x + 2 y = 5

    Evaluate the bottom of the gradient fraction at the point.

  8. Evaluate the gradient at the point

    dydx∣(1, 2)=−45\left.\frac{dy}{dx}\right|_{\left(1,\ 2\right)} = - \frac{4}{5}

    Divide the numerator value by the denominator value.

  9. Treat yy as a function of xx

    y=y(x)y=y(x)

    In an implicit relation y is not isolated, but it still depends on x.

  10. Apply the chain rule to the yy-terms

    ddxf(y)=f′(y)dydx\frac{d}{dx}f(y)=f'(y)\frac{dy}{dx}

    Every function of y picks up a factor of dy/dx when differentiated in x.

  11. Use the product rule on mixed xyxy terms

    ddx(uv)=u′v+uv′\frac{d}{dx}(uv)=u'v+uv'

    A term containing both x and y needs the product rule.

  12. The derivative of a constant is zero

    ddx(c)=0\frac{d}{dx}(c)=0

    Constant terms disappear when differentiated.

  13. Recall the power rule for xx-terms

    ddxxn=nxn−1\frac{d}{dx}x^{n}=nx^{n-1}

    Pure powers of x differentiate in the usual way.

  14. Differentiate the term x2x^{2}

    ddx(x2)=2x\frac{d}{dx}\left(x^{2}\right) = 2 x

    Differentiate this term, using the chain rule if it involves y.

  15. Select the correct derivative

    dydx=−2x−yx+2y\frac{dy}{dx} = \frac{- 2 x - y}{x + 2 y}

    This matches the correct option.

Answer
−2x−yx+2y\frac{- 2 x - y}{x + 2 y}

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