derivative of 6*6^t+2*2t — step-by-step solution

Let's perform the differentiation:

\frac{d}{d t} (6 \cdot 6^{t} + 2 \cdot 2 \cdot t)

ⓘ Multiplying powers with the same base adds the exponents:

6 \cdot 6^{t}

6^{t + 1}

■Multiplications are done before additions and subtractions:

2 \cdot 2

4

\frac{d}{d t} (6^{t + 1} + 4 t)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d t} (6^{t + 1} + 4 t)

\frac{d}{d t} (6^{t + 1}) + \frac{d}{d t} (4 t)

\frac{d}{d t} (6^{t + 1}) + \frac{d}{d t} (4 t)

You can convert the expression:

6^{t + 1}

into:

e^{(t + 1) \cdot \ln (6)}

ⅰ

\frac{d}{d t} (6^{t + 1})

\frac{d}{d t} ((t + 1) \cdot \ln (6)) \cdot e^{(t + 1) \cdot \ln (6)}

\frac{d}{d t} ((t + 1) \cdot \ln (6)) \cdot e^{(t + 1) \cdot \ln (6)} + 4

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d t} ((t + 1) \cdot \ln (6))

\ln (6) \cdot \frac{d}{d t} (t + 1)

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d t} (t + 1)

1 + 0

■As exp and ln are inverse of each other, exp(n*ln(x)) = x^n:

e^{(t + 1) \cdot \ln (6)}

6^{t + 1}

\ln (6) \cdot (1 + 0) \cdot 6^{t + 1} + 4

ⓘ Adding 0 changes nothing:

1 + 0

1

\ln (6) \cdot 6^{t + 1} + 4

▶Derivative:

6^{t + 1} \cdot \ln (6) + 4