course Mth 174 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aIf t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx.. If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1. If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx. If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx.. If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I just rounded my answers up or down, so they are a little different, but still correct. ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Once again, I just have to plug in t. y = e^(-2^2)=54.60, y = e^(-1^2)= 2.71, y = e^(-.5^2)=1.28, y = e^(0^2)=1, y = e^(.5^2)=1.28, y = e^(1^2)=2.71, y = e^(2^2)=54.60 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aIf t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx. If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx. If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx. If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1. If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx. If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx. If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z. What would be the 'chain' of simple functions for the function y = cos ( ln(x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You would first have to solve for whats inside, so z=ln(x), then y would equal cos(z). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z). Thus we have y = cos(z) = cos( ln(x) ). We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The z would be ln(t) and then y=z^2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2. Thus we have y = z^2 = (ln(t))^2. We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q005. What would be the chain of functions for y = ln ( cos(x) )? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Working from the inside, the z would be =cos(x) and then y=ln(z) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z). Thus we have y = ln(z) = ln(cos(x)). This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q006. The rule for the derivative of a chain of functions is as follows: The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ). For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be (cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) . g(x) = x^2 so g'(x) = 2 x. f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2). Thus we obtain the derivative (cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) = 2 x * ( - sin ( x^2 ) ) = - 2 x sin ( x^2). Apply the rule to find the derivative of y = sin ( ln ( x ) ) . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would use the formula f ' ( g ( x ) ) * g ' ( x )= the derivative of the f function which is cos(ln(x)) times the derivative of the g function, which is 1/x, so the answer is cos(ln(x))*1/x confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ). Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x. Since f(z) = sin(z) we have f ' (z) = cos(z). Thus the derivative of y = sin( ln (x) ) is y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) = 1 / x * cos( ln(x) ). Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q007. Find the derivative of y = ln ( 5 x^7 ) . YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula f ' ( g ( x ) ) * g ' ( x ), you can evaluate that 1/(5x^7)*35x^6 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus f ' (z) = 1 / z and g ' (x) = 35 x^6. We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7). So the derivative of y = ln( 5 x^7) is y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ]. Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q008. Find the derivative of y = e ^ ( t ^ 2 ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the formula f ' ( g ( x ) ) * g ' ( x ), we can evaluate e^(t^2)*2t confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t. Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2). Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q009. Find the derivative of y = cos ( e^t ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Once again, using the formula f ' ( g ( x ) ) * g ' ( x ), the derivative is -sin(e^t)*e^t confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t. Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t). Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function. STUDENT QUESTION: I am confused as to why the end result is - e^t sin (e^t). Is it because we just combined the multiplication?
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Given Solution: `aWe have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q011. Find the derivative of y = sin^4 ( x ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’= f ' ( g ( x ) ) * g ' ( x )= 4sin(x)^3 * cos(x) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power. We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x). STUDENT QUESTION: I am having a hard time figuring out the simplication of this one? Any help? INSTRUCTOR RESPONSE: You had the right function for the f and g functions, except that you reversed them: If f(z) = sin(z) and g(x) = x^4, then f(g(x)) = sin(x^4), and the derivative would indeed be 4 x^3 * cos(x^4). However the function here is sin^4(x), so f(z) = z^4 and g(x) = sin(x). This gives a very different result, as shown in the given solution. Regarding the simplification step cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x): a * b * c = b * a * c; this actually requires multiple applications of associative and commutative laws for addition, but it's something we're used to so we don't usually think about it that deeply. We know it is 'safe' to change the order of a sequence of multiplications. cos(x) * 4 ( sin(x) ) ^ 3 is just a sequence of multiplications. If a = cos(x), b = 4 and c = (sin(x))^3, then this expression is a * b * c. Then b * a * c = 4 cos(x) sin^3 (x). Note that sin^3(x) means (sin(x))^3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I explained how I solved for the derivative using the chain rule in question 6. I am not sure if you wanted me to do the same for the rest, but I figured them out all the same way. ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q012. Find the derivative of y = cos ( 3x ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y’= f ' ( g ( x ) ) * g ' ( x )= -sin (3x) * 3=-3sin(3x) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x). "