The tangent of x is defined as sine of x divided by cosine of x.
tanx = ˢⁱⁿˣ⁄cₒsₓ
Since tanx can be written as sinx divided by cosx, we can find the derivative of tanx using quotient rule.
Consider a function defined by y as shown below.
y = ᵘ⁄ᵥ
Let u and v be the functions of x.
Then the derivative of y with respect to x :
The above formula is called the Quotient Rule of Derivative.
Derivative of Tanx :
Substitute tanx for y, sinx for u and cosx for v into the above formula and and find the derivative of tanx.
Therefore, the derivative of tangent of x is sec^{2}x.
Find the derivative of each of the following.
Problem 1 :
tan(3x)
Solution :
We already know the derivative of tanx, which is sec^{2}x. We can find the derivative of tan(3x) using chain rule.
= [tan(3x)]'
= [sec^{2}(3x)](3x)'
= [sec^{2}(3x)](3)
= 3sec^{2}(3x)
Problem 2 :
tan(5x - 6)
Solution :
= [tan(5x - 6)]'
= [sec^{2}(5x - 6)](5x - 6)'
= [sec^{2}(5x - 6)](5)
= 5sec^{2}(5x - 6)
Problem 3 :
tan(2x^{2} - 3x + 1)
Solution :
= [tan(2x^{2} - 3x + 1)]'
= [sec^{2}(2x^{2} - 3x + 1)](2x^{2} - 3x + 1)'
= [sec^{2}(2x^{2} - 3x + 1)](4x - 3)
= (4x - 3)sec^{2}(2x^{2} - 3x + 1)
Problem 4 :
tan^{2}x
Solution :
= (tan^{2}x)'
= (2tan^{2-1}x)(tanx)'
= (2tanx)(sec^{2}x)
Problem 5 :
Solution :
Problem 6 :
tan√x
Solution :
Problem 7 :
e^{tanx}
Solution :
= (e^{tanx})'
= e^{tanx}(tanx)'
= e^{tanx}(sec^{2}x)
= (sec^{2}x)e^{tanx}
Problem 8 :
ln(tanx)
Solution :
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