Question

Differentiate the following functions: g(z)=4z^7-3z^-7+9z​

Answer 1

g(z) =

$4z {}^{7} - 3z {}^{ - 7} + 9z$

Now ,

=> g'(z) =

$7 \times 4 {z}^{6} - 3 \times ( - 7) \times {z}^{ - 7 - 1} + 9$

=

$28z {}^{6} + 21 {z}^{ - 8} + 9$

==> FORMULAS USED ARE ,

$\frac{dx {}^{n} }{dx} = n \times x {}^{n - 1}$

Answer 2

Given:

$\rm g(z) = 4 {z}^{7} - 3 {z}^{ - 7} + 9z$

To Find:

$\rm g'(z)$

Answer:

$\rm \implies \dfrac{d}{dz} (g(z)) \\ \\ \rm \implies \dfrac{d}{dz} (4 {z}^{7} - 3 {z}^{7} + 9z) \\ \\ \rm \implies\dfrac{d}{dz}(4 {z}^{7}) + \dfrac{d}{dz} ( - 3 {z}^{ - 7}) + \dfrac{d}{dz} (9z) \\ \\ \rm By \: using \: power \: rule : \dfrac{d}{dz} ( {z}^{n} ) = n {z}^{n - 1} : \\ \\ \rm \implies 4 \times 7 {z}^{(7 - 1)} - 3 \times ( - 7 {z}^{( - 7 - 1)} ) + 9 \times 1 {z}^{(1 - 1)} \\ \\ \rm \implies 28 {z}^{6} + 21 {z}^{-8 } + 9 {z}^{0} \\ \\ \rm \implies 28 {z}^{6} + 21 {z}^{-8 } + 9 \times 1 \\ \\ \rm \implies 28 {z}^{6} + 21 {z}^{-8 } + 9$

$\therefore$ $\boxed{\mathfrak{g'(z) =28 {z}^{6} + 21 {z}^{-8 } + 9}}$