Chain Rule 2
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Question 1 of 5
1. Question
Find the derivative`f(x)=1/((5-4x^5)^2)`Hint
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Correct!
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Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, remove the fraction by reciprocating the denominator`1/((5-4x^5)^2)` `=` `(5-4x^5)^(-2)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$(\color{#9a00c7}{5-4x^5})^{\color{#e65021}{-2}}$$ `x` `=` `5-4x^5` `n` `=` `-2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{-2}\cdot(\color{#004ec4}{5-4x^5})^{\color{#e65021}{(-2)}-1}\cdot\color{#00880A}{f'(5-4x^5)}$$ Substitute known values `=` $$-2(5-4x^5)^{-3}\cdot(\color{#00880A}{-20x^4})$$ Differentiate `5-4x^5` `=` $$40x^4(5-4x^5)^{-3}$$ Evaluate `=` `(40x^4)/((5-4x^5)^3)` Reciprocate `(5-4x^5)^{-3}` `y’=(40x^4)/((5-4x^5)^3)` -
Question 2 of 5
2. Question
Find the derivative`f(x)=sqrt(16-x^2)`Hint
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Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, convert the surd into an exponent.`sqrt(16-x^2)` `=` `(16-x^2)^(1/2)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$(\color{#9a00c7}{16-x^2})^{\color{#e65021}{\frac{1}{2}}}$$ `x` `=` `16-x^2` `n` `=` `1/2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{\frac{1}{2}}\cdot(\color{#004ec4}{16-x^2})^{\color{#e65021}{\frac{1}{2}}-1}\cdot\color{#00880A}{f'(16-x^2)}$$ Substitute known values `=` $$\frac{1}{2}(16-x^2)^{-\frac{1}{2}}\cdot(\color{#00880A}{-2x})$$ Differentiate `16-x^2` `=` $$-x(16-x^2)^{-\frac{1}{2}}$$ Evaluate `=` `-x/((16-x^2)^(1/2))` Reciprocate `(16-x^2)^{-\frac{1}{2}` `=` `-x/(sqrt(16-x^2))` Convert the exponent into a surd `y’=-x/(sqrt(16-x^2))` -
Question 3 of 5
3. Question
Find the derivative`f(x)=(x^4+4x)^(1/4)`Hint
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Fantastic!
Incorrect
Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$(\color{#9a00c7}{x^4+4x})^{\color{#e65021}{\frac{1}{4}}}$$ `x` `=` `x^4+4x` `n` `=` `1/4` Next, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{\frac{1}{4}}\cdot(\color{#004ec4}{x^4+4x})^{\color{#e65021}{\frac{1}{4}}-1}\cdot\color{#00880A}{f'(x^4+4x)}$$ Substitute known values `=` $$\frac{1}{4}(x^4+4x)^{-\frac{3}{4}}\cdot(\color{#00880A}{4x^3+4})$$ Differentiate `x^4+4x` `=` $$x^3+1(x^4+4x)^{-\frac{3}{4}}$$ Evaluate `=` `(x^3+1)/((x^4+4x)^(3/4))` Reciprocate `(x^4+4x)^{-\frac{3}{4}` `y’=(x^3+1)/((x^4+4x)^(3/4))` -
Question 4 of 5
4. Question
Find the derivative`f(x)=sqrt(5-6x)`Hint
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Well Done!
Incorrect
Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, convert the surd into an exponent.`sqrt(5-6x)` `=` `(5-6x)^(1/2)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$(\color{#9a00c7}{5-6x})^{\color{#e65021}{\frac{1}{2}}}$$ `x` `=` `5-6x` `n` `=` `1/2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{\frac{1}{2}}\cdot(\color{#004ec4}{5-6x})^{\color{#e65021}{\frac{1}{2}}-1}\cdot\color{#00880A}{f'(5-6x)}$$ Substitute known values `=` $$\frac{1}{2}(5-6x)^{-\frac{1}{2}}\cdot(\color{#00880A}{-6})$$ Differentiate `5-6x` `=` $$-3(5-6x)^{-\frac{1}{2}}$$ Evaluate `=` `-3/((5-6x)^(1/2))` Reciprocate `(5-6x)^{-\frac{1}{2}` `=` `-3/(sqrt(5-6x))` Convert the exponent into a surd `y’=-3/(sqrt(5-6x))` -
Question 5 of 5
5. Question
Find the derivative`f(x)=2/(sqrt(1-2x))`Hint
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Great Work!
Incorrect
Chain Rule
$$y’=\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$First, convert the surd into an exponent.`2/(sqrt(1-2x))` `=` `2/((1-2x)^(1/2))` Next, remove the fraction by reciprocating the denominator`2/((1-2x)^(1/2))` `=` `2(1-2x)^(-1/2)` Next, identify the values of the function`f(x)` `=` $$\color{#9a00c7}{x}^{\color{#e65021}{n}}$$ `f(x)` `=` $$2(\color{#9a00c7}{1-2x})^{\color{#e65021}{-\frac{1}{2}}}$$ `x` `=` `1-2x` `n` `=` `-1/2` Finally, substitute the values into the chain rule`y’` `=` $$\color{#e65021}{n}\cdot(\color{#004ec4}{f(x)})^{\color{#e65021}{n}-1}\cdot\color{#00880A}{f'(x)}$$ `=` $$\color{#e65021}{-\frac{1}{2}}\cdot2(\color{#004ec4}{1-2x})^{\color{#e65021}{-\frac{1}{2}}-1}\cdot\color{#00880A}{f'(1-2x)}$$ Substitute known values `=` $$-1(1-2x)^{-\frac{3}{2}}\cdot(\color{#00880A}{-2})$$ Differentiate `1-2x` `=` $$2(1-2x)^{-\frac{3}{2}}$$ Evaluate `=` `2/((1-2x)^(3/2))` Reciprocate `(1-2x)^{-\frac{3}{2}` `=` `2/(sqrt((1-2x)^3))` Convert the exponent into a surd `y’=2/(sqrt((1-2x)^3)`