Calculus Examples

Find the Tangent Line at (5,0) y=(x^3-25x)^14 at the point (5,0)
y=(x3-25x)14y=(x325x)14 at the point (5,0)(5,0)
Step 1
Find the first derivative and evaluate at x=5x=5 and y=0y=0 to find the slope of the tangent line.
Tap for more steps...
Step 1.1
Differentiate using the chain rule, which states that ddx[f(g(x))]ddx[f(g(x))] is f(g(x))g(x) where f(x)=x14 and g(x)=x3-25x.
Tap for more steps...
Step 1.1.1
To apply the Chain Rule, set u as x3-25x.
ddu[u14]ddx[x3-25x]
Step 1.1.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=14.
14u13ddx[x3-25x]
Step 1.1.3
Replace all occurrences of u with x3-25x.
14(x3-25x)13ddx[x3-25x]
14(x3-25x)13ddx[x3-25x]
Step 1.2
Differentiate.
Tap for more steps...
Step 1.2.1
By the Sum Rule, the derivative of x3-25x with respect to x is ddx[x3]+ddx[-25x].
14(x3-25x)13(ddx[x3]+ddx[-25x])
Step 1.2.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=3.
14(x3-25x)13(3x2+ddx[-25x])
Step 1.2.3
Since -25 is constant with respect to x, the derivative of -25x with respect to x is -25ddx[x].
14(x3-25x)13(3x2-25ddx[x])
Step 1.2.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
14(x3-25x)13(3x2-251)
Step 1.2.5
Multiply -25 by 1.
14(x3-25x)13(3x2-25)
14(x3-25x)13(3x2-25)
Step 1.3
Evaluate the derivative at x=5.
14((5)3-255)13(3(5)2-25)
Step 1.4
Simplify.
Tap for more steps...
Step 1.4.1
Simplify each term.
Tap for more steps...
Step 1.4.1.1
Raise 5 to the power of 3.
14(125-255)13(3(5)2-25)
Step 1.4.1.2
Multiply -25 by 5.
14(125-125)13(3(5)2-25)
14(125-125)13(3(5)2-25)
Step 1.4.2
Simplify the expression.
Tap for more steps...
Step 1.4.2.1
Subtract 125 from 125.
14013(3(5)2-25)
Step 1.4.2.2
Raising 0 to any positive power yields 0.
140(3(5)2-25)
Step 1.4.2.3
Multiply 14 by 0.
0(3(5)2-25)
0(3(5)2-25)
Step 1.4.3
Simplify each term.
Tap for more steps...
Step 1.4.3.1
Raise 5 to the power of 2.
0(325-25)
Step 1.4.3.2
Multiply 3 by 25.
0(75-25)
0(75-25)
Step 1.4.4
Simplify the expression.
Tap for more steps...
Step 1.4.4.1
Subtract 25 from 75.
050
Step 1.4.4.2
Multiply 0 by 50.
0
0
0
0
Step 2
Plug the slope and point values into the point-slope formula and solve for y.
Tap for more steps...
Step 2.1
Use the slope 0 and a given point (5,0) to substitute for x1 and y1 in the point-slope form y-y1=m(x-x1), which is derived from the slope equation m=y2-y1x2-x1.
y-(0)=0(x-(5))
Step 2.2
Simplify the equation and keep it in point-slope form.
y+0=0(x-5)
Step 2.3
Solve for y.
Tap for more steps...
Step 2.3.1
Add y and 0.
y=0(x-5)
Step 2.3.2
Multiply 0 by x-5.
y=0
y=0
y=0
Step 3
 [x2  12  π  xdx ]