What is the d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) ?

The d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) is a(-sin(t)+(sec^2(t))/(ln(10)tan(t)))

What is the first d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) ?

The first d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) is a(-sin(t)+(sec^2(t))/(ln(10)tan(t)))

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\frac{d}{d t} (a (cos (t) + lo g_{10} (\frac{tan ( t )}{2})))

a (- sin (t) + \frac{sec ^{2} ( t )}{ln ( 10 ) tan ( t )})

Solution steps

\frac{d}{d t} (a (cos (t) + lo g_{10} (\frac{tan ( t )}{2})))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= a \frac{d}{d t} (cos (t) + lo g_{10} (\frac{tan ( t )}{2}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= a (\frac{d}{d t} (cos (t)) + \frac{d}{d t} (lo g_{10} (\frac{tan ( t )}{2})))

\frac{d}{d t} (cos (t)) = - sin (t)

\frac{d}{d t} (lo g_{10} (\frac{tan ( t )}{2})) = \frac{sec ^{2} ( t )}{ln ( 10 ) tan ( t )}

= a (- sin (t) + \frac{sec ^{2} ( t )}{ln ( 10 ) tan ( t )})

\frac{( 1 + x )}{( 1 + x ^{2} )}

\frac{x + 1}{2 x - 1}

y = lo g_{10} (2 x)

sin (y^{2})

y = sin^{2} (x) 0.2 x

What is the d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) ?
The d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) is a(-sin(t)+(sec^2(t))/(ln(10)tan(t)))
What is the first d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) ?
The first d/(dt)(a(cos(t)+log_{10}((tan(t))/2))) is a(-sin(t)+(sec^2(t))/(ln(10)tan(t)))