Enter a problem...
Calculus Examples
10sec(x)+5tan(x)10sec(x)+5tan(x)
Step 1
Step 1.1
Find the first derivative.
Step 1.1.1
By the Sum Rule, the derivative of 10sec(x)+5tan(x)10sec(x)+5tan(x) with respect to xx is ddx[10sec(x)]+ddx[5tan(x)]ddx[10sec(x)]+ddx[5tan(x)].
ddx[10sec(x)]+ddx[5tan(x)]ddx[10sec(x)]+ddx[5tan(x)]
Step 1.1.2
Evaluate ddx[10sec(x)]ddx[10sec(x)].
Step 1.1.2.1
Since 1010 is constant with respect to xx, the derivative of 10sec(x)10sec(x) with respect to xx is 10ddx[sec(x)]10ddx[sec(x)].
10ddx[sec(x)]+ddx[5tan(x)]10ddx[sec(x)]+ddx[5tan(x)]
Step 1.1.2.2
The derivative of sec(x)sec(x) with respect to xx is sec(x)tan(x)sec(x)tan(x).
10sec(x)tan(x)+ddx[5tan(x)]10sec(x)tan(x)+ddx[5tan(x)]
10sec(x)tan(x)+ddx[5tan(x)]10sec(x)tan(x)+ddx[5tan(x)]
Step 1.1.3
Evaluate ddx[5tan(x)]ddx[5tan(x)].
Step 1.1.3.1
Since 55 is constant with respect to xx, the derivative of 5tan(x)5tan(x) with respect to xx is 5ddx[tan(x)]5ddx[tan(x)].
10sec(x)tan(x)+5ddx[tan(x)]10sec(x)tan(x)+5ddx[tan(x)]
Step 1.1.3.2
The derivative of tan(x)tan(x) with respect to xx is sec2(x)sec2(x).
f′(x)=10sec(x)tan(x)+5sec2(x)
f′(x)=10sec(x)tan(x)+5sec2(x)
f′(x)=10sec(x)tan(x)+5sec2(x)
Step 1.2
The first derivative of f(x) with respect to x is 10sec(x)tan(x)+5sec2(x).
10sec(x)tan(x)+5sec2(x)
10sec(x)tan(x)+5sec2(x)
Step 2
Step 2.1
Set the first derivative equal to 0.
10sec(x)tan(x)+5sec2(x)=0
Step 2.2
Graph each side of the equation. The solution is the x-value of the point of intersection.
x=7π6+2πn,11π6+2πn, for any integer n
x=7π6+2πn,11π6+2πn, for any integer n
Step 3
Step 3.1
Set the argument in sec(x) equal to π2+πn to find where the expression is undefined.
x=π2+πn, for any integer n
Step 3.2
The equation is undefined where the denominator equals 0, the argument of a square root is less than 0, or the argument of a logarithm is less than or equal to 0.
{x|x=π2+πn}n, for any integer n
{x|x=π2+πn}n, for any integer n
Step 4
Step 4.1
Evaluate at x=7π6.
Step 4.1.1
Substitute 7π6 for x.
10sec(7π6)+5tan(7π6)
Step 4.1.2
Simplify.
Step 4.1.2.1
Simplify each term.
Step 4.1.2.1.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the third quadrant.
10(-sec(π6))+5tan(7π6)
Step 4.1.2.1.2
The exact value of sec(π6) is 2√3.
10(-2√3)+5tan(7π6)
Step 4.1.2.1.3
Multiply 2√3 by √3√3.
10(-(2√3⋅√3√3))+5tan(7π6)
Step 4.1.2.1.4
Combine and simplify the denominator.
Step 4.1.2.1.4.1
Multiply 2√3 by √3√3.
10(-2√3√3√3)+5tan(7π6)
Step 4.1.2.1.4.2
Raise √3 to the power of 1.
10(-2√3√31√3)+5tan(7π6)
Step 4.1.2.1.4.3
Raise √3 to the power of 1.
10(-2√3√31√31)+5tan(7π6)
Step 4.1.2.1.4.4
Use the power rule aman=am+n to combine exponents.
10(-2√3√31+1)+5tan(7π6)
Step 4.1.2.1.4.5
Add 1 and 1.
10(-2√3√32)+5tan(7π6)
Step 4.1.2.1.4.6
Rewrite √32 as 3.
Step 4.1.2.1.4.6.1
Use n√ax=axn to rewrite √3 as 312.
10(-2√3(312)2)+5tan(7π6)
Step 4.1.2.1.4.6.2
Apply the power rule and multiply exponents, (am)n=amn.
10(-2√3312⋅2)+5tan(7π6)
Step 4.1.2.1.4.6.3
Combine 12 and 2.
10(-2√3322)+5tan(7π6)
Step 4.1.2.1.4.6.4
Cancel the common factor of 2.
Step 4.1.2.1.4.6.4.1
Cancel the common factor.
10(-2√3322)+5tan(7π6)
Step 4.1.2.1.4.6.4.2
Rewrite the expression.
10(-2√331)+5tan(7π6)
10(-2√331)+5tan(7π6)
Step 4.1.2.1.4.6.5
Evaluate the exponent.
10(-2√33)+5tan(7π6)
10(-2√33)+5tan(7π6)
10(-2√33)+5tan(7π6)
Step 4.1.2.1.5
Multiply 10(-2√33).
Step 4.1.2.1.5.1
Multiply -1 by 10.
-102√33+5tan(7π6)
Step 4.1.2.1.5.2
Combine -10 and 2√33.
-10(2√3)3+5tan(7π6)
Step 4.1.2.1.5.3
Multiply 2 by -10.
-20√33+5tan(7π6)
-20√33+5tan(7π6)
Step 4.1.2.1.6
Move the negative in front of the fraction.
-20√33+5tan(7π6)
Step 4.1.2.1.7
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
-20√33+5tan(π6)
Step 4.1.2.1.8
The exact value of tan(π6) is √33.
-20√33+5√33
Step 4.1.2.1.9
Combine 5 and √33.
-20√33+5√33
-20√33+5√33
Step 4.1.2.2
Simplify terms.
Step 4.1.2.2.1
Combine the numerators over the common denominator.
-20√3+5√33
Step 4.1.2.2.2
Add -20√3 and 5√3.
-15√33
Step 4.1.2.2.3
Cancel the common factor of -15 and 3.
Step 4.1.2.2.3.1
Factor 3 out of -15√3.
3(-5√3)3
Step 4.1.2.2.3.2
Cancel the common factors.
Step 4.1.2.2.3.2.1
Factor 3 out of 3.
3(-5√3)3(1)
Step 4.1.2.2.3.2.2
Cancel the common factor.
3(-5√3)3⋅1
Step 4.1.2.2.3.2.3
Rewrite the expression.
-5√31
Step 4.1.2.2.3.2.4
Divide -5√3 by 1.
-5√3
-5√3
-5√3
-5√3
-5√3
-5√3
Step 4.2
Evaluate at x=11π6.
Step 4.2.1
Substitute 11π6 for x.
10sec(11π6)+5tan(11π6)
Step 4.2.2
Simplify.
Step 4.2.2.1
Simplify each term.
Step 4.2.2.1.1
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.
10sec(π6)+5tan(11π6)
Step 4.2.2.1.2
The exact value of sec(π6) is 2√3.
102√3+5tan(11π6)
Step 4.2.2.1.3
Multiply 2√3 by √3√3.
10(2√3⋅√3√3)+5tan(11π6)
Step 4.2.2.1.4
Combine and simplify the denominator.
Step 4.2.2.1.4.1
Multiply 2√3 by √3√3.
102√3√3√3+5tan(11π6)
Step 4.2.2.1.4.2
Raise √3 to the power of 1.
102√3√31√3+5tan(11π6)
Step 4.2.2.1.4.3
Raise √3 to the power of 1.
102√3√31√31+5tan(11π6)
Step 4.2.2.1.4.4
Use the power rule aman=am+n to combine exponents.
102√3√31+1+5tan(11π6)
Step 4.2.2.1.4.5
Add 1 and 1.
102√3√32+5tan(11π6)
Step 4.2.2.1.4.6
Rewrite √32 as 3.
Step 4.2.2.1.4.6.1
Use n√ax=axn to rewrite √3 as 312.
102√3(312)2+5tan(11π6)
Step 4.2.2.1.4.6.2
Apply the power rule and multiply exponents, (am)n=amn.
102√3312⋅2+5tan(11π6)
Step 4.2.2.1.4.6.3
Combine 12 and 2.
102√3322+5tan(11π6)
Step 4.2.2.1.4.6.4
Cancel the common factor of 2.
Step 4.2.2.1.4.6.4.1
Cancel the common factor.
102√3322+5tan(11π6)
Step 4.2.2.1.4.6.4.2
Rewrite the expression.
102√331+5tan(11π6)
102√331+5tan(11π6)
Step 4.2.2.1.4.6.5
Evaluate the exponent.
102√33+5tan(11π6)
102√33+5tan(11π6)
102√33+5tan(11π6)
Step 4.2.2.1.5
Multiply 102√33.
Step 4.2.2.1.5.1
Combine 10 and 2√33.
10(2√3)3+5tan(11π6)
Step 4.2.2.1.5.2
Multiply 2 by 10.
20√33+5tan(11π6)
20√33+5tan(11π6)
Step 4.2.2.1.6
Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the fourth quadrant.
20√33+5(-tan(π6))
Step 4.2.2.1.7
The exact value of tan(π6) is √33.
20√33+5(-√33)
Step 4.2.2.1.8
Multiply 5(-√33).
Step 4.2.2.1.8.1
Multiply -1 by 5.
20√33-5√33
Step 4.2.2.1.8.2
Combine -5 and √33.
20√33+-5√33
20√33+-5√33
Step 4.2.2.1.9
Move the negative in front of the fraction.
20√33-5√33
20√33-5√33
Step 4.2.2.2
Simplify terms.
Step 4.2.2.2.1
Combine the numerators over the common denominator.
20√3-5√33
Step 4.2.2.2.2
Subtract 5√3 from 20√3.
15√33
Step 4.2.2.2.3
Cancel the common factor of 15 and 3.
Step 4.2.2.2.3.1
Factor 3 out of 15√3.
3(5√3)3
Step 4.2.2.2.3.2
Cancel the common factors.
Step 4.2.2.2.3.2.1
Factor 3 out of 3.
3(5√3)3(1)
Step 4.2.2.2.3.2.2
Cancel the common factor.
3(5√3)3⋅1
Step 4.2.2.2.3.2.3
Rewrite the expression.
5√31
Step 4.2.2.2.3.2.4
Divide 5√3 by 1.
5√3
5√3
5√3
5√3
5√3
5√3
Step 4.3
List all of the points.
(7π6+2πn,-5√3),(11π6+2πn,5√3), for any integer n
(7π6+2πn,-5√3),(11π6+2πn,5√3), for any integer n
Step 5
