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MathTake the derivative of `y=x^3+x` . `y'= 3x^2+1` For horizontal tangent lines, the slope is zero. Set the derivative function to zero because we want to know where x is when the slope is zero....

MathRewrite the equation and take the derivative. Use the power rule. `y= 1/x^2 = x^(2)` `y' = 2x^(3)` `y' = 2/x^3` For horizontal tangent lines, the slope will always equal zero. Set the...

MathSo I will start by rewriting this equation in the form: `f(x)=x^2+9 ` We know that if we have `f'(x)` then we can plug in any value for x, and the result will be what the slope of the function is...

MathWe will need to take the derivative of the function and set the derivative equal to zero. `y' = 1+cos(x)` `0 = 1+cos(x)` `cos(x)= 1` The value of `x` is the angle where we have a point on the unit...

MathTake the derivative of the function. `y' = sqrt3 2sin(x)` Set the derivative function equal to zero, since we want to know where x is when the slope is zero. `0 = sqrt3 2sin(x)` `2sin(x)=sqrt3`...

MathThe tangent line will touch a point on the function f(x). Set both equations equal to each other. `kx^2= 6x+1` With two unknown variables, we will need another relationship of x and k. Take the...

MathThe tangent line will touch a point on the original function f(x). Set the two equations equal to each other since they intersect. `kx^2=2x+3` We will need another relationship since we have two...

MathThe original equation will intersect a point on the tangent line. Set both equations equal to each other. `k/x = 3/4 x +3` We have 2 unknowns and 1 equation. Take the derivative of f(x) in order...

MathThe tangent line will touch one point of the original function. Set the two equations equal to each other. Rewrite the square root as a fractional power. `kx^(1/2)= x+4` We need a relationship of...

MathThe tangent line must touch a point on the f(x) function. Set both equations equal to each other. `kx^3 = x+1` Take the derivative of f(x) to find another relationship with k and x. The k is a...

MathThe tangent line will touch one point of the original function. This means that: `kx^4 = 4x1` We have two variables that we don't know. Find the derivative of `f(x)` . The k is a constant and...

Mathg(t) = 2*cost + 5 slope of the curve = g'(t) = d{g(t)}/dt = 2*sint Thus, slope at (pi,7) g'(pi) = 2*sin(pi) =0 Thus, the curve has a tangent parallel to the xaxis(horizontal axis) at (pi,7)

MathThe derivative of the first term can be solved using the power rule. `d/dx x^n = nx^(n1)` The derivative of `x^2` is `2x` . The derivative of a constant is always zero. Use the power rule again...

MathUse the power rule for each term. Derive each term. `d/dx x^n= nx^(n1)` `f'(x) = 3x^2 2 9x^4` Eliminate all negative exponents. The answer is: `f'(x) = 3x^2 2 9/x^4`

MathWe can rewrite g(t) so that we can avoid the quotient rule altogether. `g(t)=t^24t^3` Use the power rule to derive g(t). The power rule is: `d/dx x^n = nx^(n1)` `g'(t)= 2t +12 t^4` The answer...

MathDerive term by term. The derivative is the slope of a function. The derivative of `8x` is 8. Rewrite `3/x^2` with a negative exponent. `3/x^2= 3x^2` Use the power rule to derive. `2(3)x^3...

MathUse the quotient rule to solve this question. The formula is: `f'(x) = (BT'  TB')/ B^2` Where B and T are bottom function and top function, respectively. `B= x` `B'=1` `T=4x^3+3x^2` `T'=...

MathDivide each term by `x^3` . `f(x) = 2xx^(2)` Use the constant rule and the power rule to derive this function. `f'(x) = 2+2x^3` Eliminate the negative exponent by rewriting it as a fraction. The...

MathDivide each term by to avoid the quotient rule. `f(x) = x3+4x^2` Take the derivative of each term. Use the power rule for the last term. `f'(x)= 10+(2)(4)(x^3)` `f'(x) = 18x^3` Rewrite the...

MathDivide all terms by x. `f(x)=4x^2+2+5/x ` `f(x)= 4x^2+2+5x^1` Take the derivative of each term. `f'(x) = 8x+0 +(1)(5)(x^2)` Simplify. The answer is: `f'(x) = 8x5/x^2` ` `

MathGiven: ` y=x(x^2+1)` `y=x^3+x` `y'=3x^2+1` ``

MathGiven: `y=x^2(2x^23x)` Distribute in the x^2 to rewrite the expression as a polynomial. `y=2x^43x^3` Find the derivative. `y'=8x^39x^2`

MathRewrite the function with exponents. `f(x) = x^(1/2)6x^(1/3)` Use the power rule to solve. `f'(x) = 1/2 x^(1/2)(1/3)6x^(2/3)` Simplify and eliminate the negative fractions by writing them as a...

MathGiven: `f(t)=t^(2/3)t^(1/3)+4` The derivative is: `f'(t)=(2)/(3)t^(1/3)(1)/(3)t^(2/3)` `f'(t)=(2)/(3root(3)(t))(1)/(3root(3)(t^2))`

MathRewrite the function: `f(x)=6x^(1/2)+5cos(x)` Take the derivative of the first term by using the power rule and the derivative of the second term. `f'(x)= 1/2 (6)x^(1/2)5sin(x)` Simplify the...

MathThe first term is the same as: `2/x^(1/3) =2x^(1/3)` Take the derivative of this term. `1/3(2)x^(1/31) = 2/3x^(4/3)` The derivative of `3cos(x)` is `3sin(x)` . Combine the terms....

MathGiven: Take the derivative of this function. `y' = 4x^36x` Substitute the value of x=1 from the given point into the derivative function. `y' = 4(1)^36(1)` `y'=46 = 2` The slope at the...

Matha) So we are given: `y=x^3  3x` If we derive this function, then we will have the following: `y'=3x^2 3` This will show us what the slope of the original line would be at any x value that we plug...

MathEliminate the radical by rewriting it as a fraction. The function becomes: `f(x) = 2/ (x^3)^(1/4) = 2/x^(3/4) = 2x^(3/4)` Take the derivative by using the power rule. `f'(x) = 3/4...

MathGiven: `f(t)=2t^2+3t6` The derivative is: `f'(t)=4t+3` ``

MathGiven: `y=t^23t+1` The derivative is: `y'=2t3` ``

MathGiven: `g(x)=x^2+4x^3` The derivative is: `g'(x)=2x+12x^2` ``

MathGiven: `y=4x3x^3` The derivative is: `y'=49x^2` ``

MathGiven: `s(t)=t^3+5t^23t+8` The derivative is: `s'(t)=3t^2+10t3` ``

MathGiven: `y=2x^3+6x^21` `` The derivative is: `y'=6x^2+12x`

MathGiven: `y=(pi)/(2)sin(theta)cos(theta)` The derivative is: `y'=(pi)/(2)cos(theta)+sin(theta)` ``

MathGiven: `g(t)=picos(t)` Find the derivative using the product rule: `d/dx[f(x)g(x)]=f(x)g'(x)+g(x)f'(x)` The derivative is: `g'(t)=pi(sin(t))+cos(t)(0)` ` g'(t)=pisin(t)` ``

MathGiven: `y=x^2(1)/(2)cos(x)` The derivative is: `y'=2x+(1)/(2)sin(x)` ``

MathGiven: `y=7+sin(x)` The derivative is: `y'=cos(x)` ``

MathRewrite the `1/x` term as a fraction. `y= x^1 3sin(x)` Take the derivative of the first term using the power rule. `y'=x^2 3cos(x)` Simplify and eliminate the negative exponent. The answer...

MathGiven: `y=(5)/(2x)^3+2cos(x)` `` Simplify the given. `y=(5)/(8x^3)+2cos(x)` `y=(5x^3)/(8)+2cos(x)` Find the derivative. `y'=(15x^4)/(8)2sin(x)` `y'=(15)/(8x^4)2sin(x)`

MathRewrite the function so that we can use the power rule to take the derivative. `f(x)= 8x^2` Take the derivative. `f'(x)= 2(8) x^3` `f'(x)= 16/x^3` Substitute the given x value from the...

MathRewrite the given function so that we can proceed with the power rule instead of the quotient rule. `f(t)=24t^(1)` Take the derivative of the given function. The derivative of a constant is...

MathTake the derivative for the derivative function. `f'(x)= 21/5 x^2` Substitute the point where x=0 to find the slope at that point. `f'(x)= 21/5 (0)^2= 0` The slope at that point is zero....

MathTake the derivative of the function. `y' = 8x^3` Substitute the x value of the given point. `y'(1) = 8(1)^3 = 8` The derivative, or slope, at point (1,1) is eight. Check by graphing the...

MathUse the power rule and the chain rule to find the derivative. `y' = 2(4x+1) * (4)` `y'=8(4x+1)` `y'=32x+8` Substitute the value at x=0 from the given point. `y'=32(0)+8= 8` The slope at the...

MathTake the derivative of the given function. The chain rule will be shown in the steps. The derivative of the inner function (x4) with the respect to x is 1. `f'(x) =2 * 2(x4)^(21) * (1)`...

MathTake the derivative. Be careful that the derivative of theta with respect to theta is one! `f'(theta)= 4cos(theta)1` Substitute the point given to determine the derivative: ` ` `4cos(0)1=...

MathGiven: `y=12` The derivative of a constant will always be zero: The derivative is: `y'=0` ``

MathGiven: `f(x)=9` The derivative of a constant will always be zero. The derivative is: `f'(x)=0` ``