Question

prove that differentiation of tanx is sec2x

Answer 1

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Answers 

Let y = tan(x)
tan(x) as sin(x)/cos(x)

Therefore y = sin(x)/cos(x)

Use the quotient rule, which states that for y = f(x)/g(x), dy/dx = (f'(x)g(x) - f(x)g'(x))/g2(x) with f(x) = sin(x) and g(x) = cos(x).

Recall the derivatives of sin(x) as cos(x) and cos(x) as -sinx

dy/dx = (cos(x)*cos(x) + sin(x)*sin(x)) / cos2x
the trigonometric identity sin2(x) + cos2(x) = 1

Therefore dy/dx = 1/cos2(x) = sec2(x)

Answer 2

wait a min,
let y = tanx

recall the tanx = sinx/cosx

therefore y= sinx / cosx


now use the quotient rule , which states that for y = f(x)/g(x)

dy/dx= f'(x) . g(x) - g'(x) . f(x) / g^2(x)

f(x)= sinx and g(x) = cosx

now this gives us

dy/dx = cos(x)*. cos(x) + sin (x)* . sin(x) / cos^2x

now the trigonometric identity is sin^2(x) + cos ^2(x) = 1

therefore dy/dx = 1/cos^2(x) = sec^2x