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Title: Calculus: Differentiation using the chain rule. Target: On completion of this worksheet you should be able to use the chain rule to differentiate functions of a function. Chain Rule of Differentiation in Calculus. The chain rule of differentiation of functions in calculus is presented along with several examples and detailed solutions and comments. Also in this site, Step by Step Calculator to Find Derivatives Using Chain Rule.
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3.3 Chain Ruleap Calculus Calculator
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Find the Derivative f(x)=x(x-4)^3. Differentiate using the chain rule, which states that is where. Tap for more steps. To apply the Chain Rule. Unit 3: Trigonometry and the Chain Rule The chain rule is the most complex of the differentiation procedures. It is also the most powerful: with it you will be able to differentiate any algebraic function. By a trick of notation, it looks innocuous enough: Just cancel the du s. It should be stressed that this is a trick of the notation; the. This lesson contains the following Essential Knowledge (EK) concepts for the.AP Calculus course.Click here for an overview of all the EK's in this course. EK 2.1C4. AP® is a trademark registered and owned by the College Board, which was not involved in the production of, and does not endorse, this site.
Section 3-9 : Chain Rule
For problems 1 – 27 differentiate the given function.
3.3 Chain Ruleap Calculus 14th Edition
- (fleft( x right) = {left( {6{x^2} + 7x} right)^4}) Solution
- (gleft( t right) = {left( {4{t^2} - 3t + 2} right)^{ - 2}}) Solution
- (y = sqrt[3]{{1 - 8z}}) Solution
- (Rleft( w right) = csc left( {7w} right)) Solution
- (Gleft( x right) = 2sin left( {3x + tan left( x right)} right)) Solution
- (hleft( u right) = tan left( {4 + 10u} right)) Solution
- (fleft( t right) = 5 + {{bf{e}}^{4t + {t^{,7}}}}) Solution
- (gleft( x right) = {{bf{e}}^{1 - cos left( x right)}}) Solution
- (Hleft( z right) = {2^{1 - 6z}}) Solution
- (uleft( t right) = {tan ^{ - 1}}left( {3t - 1} right)) Solution
- (Fleft( y right) = ln left( {1 - 5{y^2} + {y^3}} right)) Solution
- (Vleft( x right) = ln left( {sin left( x right) - cot left( x right)} right)) Solution
- (hleft( z right) = sin left( {{z^6}} right) + {sin ^6}left( z right)) Solution
- (Sleft( w right) = sqrt {7w} + {{bf{e}}^{ - w}}) Solution
- (gleft( z right) = 3{z^7} - sin left( {{z^2} + 6} right)) Solution
- (fleft( x right) = ln left( {sin left( x right)} right) - {left( {{x^4} - 3x} right)^{10}}) Solution
- (hleft( t right) = {t^6},sqrt {5{t^2} - t} ) Solution
- (qleft( t right) = {t^2}ln left( {{t^5}} right)) Solution
- (gleft( w right) = cos left( {3w} right)sec left( {1 - w} right)) Solution
- (displaystyle y = frac{{sin left( {3t} right)}}{{1 + {t^2}}}) Solution
- (displaystyle Kleft( x right) = frac{{1 + {{bf{e}}^{ - 2x}}}}{{x + tan left( {12x} right)}}) Solution
- (fleft( x right) = cos left( {{x^2}{{bf{e}}^x}} right)) Solution
- (z = sqrt {5x + tan left( {4x} right)} ) Solution
- (fleft( t right) = {left( {{{bf{e}}^{ - 6t}} + sin left( {2 - t} right)} right)^3}) Solution
- (gleft( x right) = {left( {ln left( {{x^2} + 1} right) - {{tan }^{ - 1}}left( {6x} right)} right)^{10}}) Solution
- (hleft( z right) = {tan ^4}left( {{z^2} + 1} right)) Solution
- (fleft( x right) = {left( {sqrt[3]{{12x}} + {{sin }^2}left( {3x} right)} right)^{ - 1}}) Solution
- Find the tangent line to (fleft( x right) = 4sqrt {2x} - 6{{bf{e}}^{2 - x}}) at (x = 2). Solution
- Determine where (Vleft( z right) = {z^4}{left( {2z - 8} right)^3}) is increasing and decreasing. Solution
- The position of an object is given by (sleft( t right) = sin left( {3t} right) - 2t + 4). Determine where in the interval (left[ {0,3} right]) the object is moving to the right and moving to the left. Solution
- Determine where (Aleft( t right) = {t^2}{{bf{e}}^{5 - t}}) is increasing and decreasing. Solution
- Determine where in the interval (left[ { - 1,20} right]) the function (fleft( x right) = ln left( {{x^4} + 20{x^3} + 100} right)) is increasing and decreasing. Solution