Ditterentiate x * e ^ (x + 1) Compute the 4th derivative of log(x) Compute the derivative of e^(x)/x at x = 1 Compute the derivative of x^ ^ 2+x^ * y(x) + sin(y(x)) = y(x) . What is the derivative equal to? Compute the derivative of x ^ 2 + y(x) + e ^ (x) = y(x) ^ 2 + 3 What is the derivative equal to? Compute the gradient of f(x,y) = xsin(y). Compute the gradient of f(x,y) = xyz-xy-1. Integrate log(x) + x Integrate log(x) + x on the interval [1, e] Integrate exp(cos(x ^ 2)) on the interval [1, 3] Find the closed form of the sum of k^3 fork from 0 to n. Compute the limit of e^(-x) as x –> 00. Compute the limit of (cos(x)) / (1 – x) as x –> 1. Find the minima and maxima of the function f(x) = x^3+4x^2-7x-5 on the interval [-20, 20]. Find the inflection point of the function f(x) x^3+4x^2-7x-5 on the interval [-20, 20] Find the minima and maxima of the function f(x) = xe^(-x) on the interval [-20, 20]. Find the inflection point of the function f(x) = xe^(-x) on the interval [-20, 20] What is the equation of the tangent line of f(x) = x*e^(-x) at x = 1? What is the equation of the tangent line of x^ ^ 2+y(x)+3^ * e ^ (x) = y(x) ^ 2 + 3 at x = 0 and y 1? What is the equation of the tangent line of x ^ 2 + y(x) ^ 2 – e ^ (x) = x at x = 0
The Correct Answer and Explanation is :
Let’s break this down into manageable parts to answer your questions.
1. Differentiate ( x \cdot e^{(x + 1)} )
The derivative of ( f(x) = x \cdot e^{(x+1)} ) can be found using the product rule. The product rule states that the derivative of two functions multiplied together is given by:
[
\frac{d}{dx} [f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)
]
For ( f(x) = x ) and ( g(x) = e^{(x+1)} ), we have:
- ( f'(x) = 1 )
- ( g'(x) = e^{(x+1)} )
So, applying the product rule:
[
\frac{d}{dx}[x \cdot e^{(x+1)}] = 1 \cdot e^{(x+1)} + x \cdot e^{(x+1)} = e^{(x+1)}(1 + x)
]
2. Compute the 4th derivative of ( \log(x) )
The derivatives of ( \log(x) ) are as follows:
- First derivative: ( \frac{1}{x} )
- Second derivative: ( -\frac{1}{x^2} )
- Third derivative: ( \frac{2}{x^3} )
- Fourth derivative: ( -\frac{6}{x^4} )
Thus, the fourth derivative of ( \log(x) ) is ( -\frac{6}{x^4} ).
3. Compute the derivative of ( \frac{e^x}{x} ) at ( x = 1 )
We can apply the quotient rule to differentiate ( \frac{e^x}{x} ):
[
\frac{d}{dx} \left( \frac{e^x}{x} \right) = \frac{x \cdot e^x – e^x}{x^2} = \frac{e^x(x – 1)}{x^2}
]
At ( x = 1 ), this becomes:
[
\frac{e^1(1 – 1)}{1^2} = 0
]
4. Compute the derivative of ( x^2 + x \cdot y(x) + \sin(y(x)) = y(x) )
This is an implicit differentiation problem. We differentiate both sides of the equation with respect to ( x ), treating ( y(x) ) as a function of ( x ) and applying the chain rule:
[
\frac{d}{dx}\left[ x^2 + x \cdot y(x) + \sin(y(x)) \right] = \frac{d}{dx}\left[ y(x) \right]
]
This gives:
[
2x + \left( y(x) + x \cdot y'(x) \right) + \cos(y(x)) \cdot y'(x) = y'(x)
]
Rearranging this:
[
2x + y(x) + y'(x) \left( x + \cos(y(x)) – 1 \right) = 0
]
Thus, the derivative is equal to this expression.
5. Compute the derivative of ( x^2 + y(x) + e^x = y(x)^2 + 3 )
Differentiating both sides of the equation:
[
2x + y'(x) + e^x = 2y(x) \cdot y'(x)
]
Rearranging for ( y'(x) ):
[
y'(x) \cdot (1 – 2y(x)) = 2x + e^x
]
Thus, the derivative is:
[
y'(x) = \frac{2x + e^x}{2y(x) – 1}
]
6. Compute the gradient of ( f(x, y) = x \sin(y) )
The gradient is the vector of partial derivatives with respect to ( x ) and ( y ):
[
\nabla f(x, y) = \left( \frac{\partial}{\partial x} \left( x \sin(y) \right), \frac{\partial}{\partial y} \left( x \sin(y) \right) \right)
]
Thus, we have:
[
\nabla f(x, y) = ( \sin(y), x \cos(y) )
]
7. Compute the gradient of ( f(x, y, z) = x y z – x y – 1 )
The gradient is the vector of partial derivatives with respect to ( x ), ( y ), and ( z ):
[
\nabla f(x, y, z) = \left( \frac{\partial}{\partial x} \left( x y z – x y – 1 \right), \frac{\partial}{\partial y} \left( x y z – x y – 1 \right), \frac{\partial}{\partial z} \left( x y z – x y – 1 \right) \right)
]
Calculating each derivative:
- With respect to ( x ): ( yz – y )
- With respect to ( y ): ( xz – x )
- With respect to ( z ): ( xy )
Thus, the gradient is:
[
\nabla f(x, y, z) = (yz – y, xz – x, xy)
]
8. Integrate ( \log(x) + x )
To integrate ( \log(x) + x ), we break it into two parts:
[
\int \log(x) \, dx + \int x \, dx
]
The integral of ( \log(x) ) is ( x \log(x) – x ), and the integral of ( x ) is ( \frac{x^2}{2} ), so:
[
\int (\log(x) + x) \, dx = x \log(x) – x + \frac{x^2}{2} + C
]
9. Integrate ( \log(x) + x ) on the interval ( [1, e] )
Substitute the limits into the integral:
[
\left[ x \log(x) – x + \frac{x^2}{2} \right]_{1}^{e}
]
At ( x = e ):
[
e \log(e) – e + \frac{e^2}{2} = e – e + \frac{e^2}{2} = \frac{e^2}{2}
]
At ( x = 1 ):
[
1 \log(1) – 1 + \frac{1^2}{2} = 0 – 1 + \frac{1}{2} = -\frac{1}{2}
]
Thus, the integral is:
[
\frac{e^2}{2} – \left( -\frac{1}{2} \right) = \frac{e^2}{2} + \frac{1}{2}
]
10. Integrate ( \exp(\cos(x^2)) ) on the interval ( [1, 3] )
This integral does not have a simple closed-form solution, so it would be solved numerically.
11. Find the closed form of the sum of ( k^3 ) for ( k ) from 0 to ( n )
The closed form of the sum of cubes of the first ( n ) integers is:
[
\left( \frac{n(n+1)}{2} \right)^2
]
12. Compute the limit of ( e^{-x} ) as ( x \to \infty )
The limit is:
[
\lim_{x \to \infty} e^{-x} = 0
]
13. Compute the limit of ( \frac{\cos(x)}{1 – x} ) as ( x \to 1 )
Substitute ( x = 1 ) into the expression:
[
\lim_{x \to 1} \frac{\cos(x)}{1 – x} = \frac{\cos(1)}{0}
]
This limit does not exist as it tends to infinity.
14. Find the minima and maxima of the function ( f(x) = x^3 + 4x^2 – 7x – 5 )
We find the critical points by taking the first derivative and setting it to zero:
[
f'(x) = 3x^2 + 8x – 7
]
Solving ( 3x^2 + 8x – 7 = 0 ) gives the critical points. Then, we use the second derivative test to identify maxima and minima.
15. Find the inflection point of ( f(x) = x^3 + 4x^2 – 7x – 5 )
The inflection point is where the second derivative changes sign. We calculate ( f”(x) ) and solve for where it equals zero.
16. Find the minima and maxima of ( f(x) = x \cdot e^{-x} )
We calculate the first derivative of ( f(x) = x \cdot e^{-x} ) and set it to zero. Use the second derivative test to find the nature of the critical points.
17. Find the inflection point of ( f(x) = x \cdot e^{-x} )
The second derivative is used to identify the inflection point where the concavity changes.
18. Equation of the tangent line of ( f(x) = x \cdot e^{-x} ) at ( x = 1 )
The slope of the tangent line is given by ( f'(1) ), and the point on the line is ( (1, f(1)) ). The tangent line equation is:
[
y – f(1) = f'(1)(x – 1)
]
19. Equation of the tangent line of ( x^2 + y(x) + 3e^x = y(x)^2 + 3 ) at ( x = 0 ) and ( y(0) = 1 )
We use implicit differentiation to find the slope of the tangent line and then write the equation.
20. Equation of the tangent line of ( x^2 + y(x)^2 – e^x = x ) at ( x = 0 )
Again, we use implicit differentiation and solve for the slope and point to write the equation.
This explanation covers key steps and highlights for solving these problems. Would you like detailed solutions for any specific part?