Question

I = d( t² +4t+ 5)/dt
=> I = 2t + 4 how it has been done

Answer 1

Solution:

We have to show: I = 2t + 4

Given That:

$\rm \longrightarrow I = \dfrac{d}{dt}( {t}^{2} + 4t + 5)$

We know that:

$\rm \longrightarrow \dfrac{d}{dt}(f + g) = \dfrac{df}{dt} + \dfrac{dg}{dt}$

Using this result, we get:

$\rm \longrightarrow I = \dfrac{d}{dt}( {t}^{2})+ \dfrac{d}{dt}(4t)+ \dfrac{d}{dt} (5)$

The derivative of a constant is always 0. 5 is constant with respect to t. So:

$\rm \longrightarrow I = \dfrac{d}{dt}( {t}^{2})+ \dfrac{d}{dt}(4t)$

We know that:

$\rm \longrightarrow \dfrac{d}{dt}( C \times {f}) =C \dfrac{d}{dt}(f)$

$\rm \longrightarrow \dfrac{d}{dt}( {t}^{n} ) =n {t}^{n - 1}$

Therefore:

$\rm \longrightarrow I = 2 {t}^{2 - 1} + 4\dfrac{d}{dt}(t)$

$\rm \longrightarrow I = 2t+ 4\dfrac{dt}{dt}$

$\rm \longrightarrow I = 2t+ 4$

Hence Proved..!!

Additional Information:

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$