# 3.3 Chain Ruleap Calculus

- 3.3 Chain Ruleap Calculus 2nd Edition
- 3.3 Chain Ruleap Calculus 14th Edition
- 3.3 Chain Ruleap Calculus Calculator
- 3.3 Chain Ruleap Calculus Solver

### 3.3 Chain Ruleap Calculus 2nd Edition

Show All NotesHide All Notes### 3.3 Chain Ruleap Calculus 14th Edition

This video explains about Topic 3.3 for subject BUM2123 Applied Calculus which will be used in Substitute Blended Learning. 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.

*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

. 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.® is a trademark registered. Chain Rule appears everywhere in the world of differential calculus. Whenever we are finding the derivative of a function, be it a composite function or not, we are in fact using the Chain Rule. In this section, we will learn about the concept, the definition and the application of the Chain Rule, as well as a secret trick – 'The Bracket. More Lessons for Calculus Math Worksheets The Chain Rule The following figure gives the Chain Rule that is used to find the derivative of composite functions. Scroll down the page for more examples and solutions. In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, then.

### 3.3 Chain Ruleap Calculus Calculator

### Section 3-9 : Chain Rule

### 3.3 Chain Ruleap Calculus Solver

For problems 1 – 27 differentiate the given function.

- (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