Find $\frac { d } { d x } \left( \sqrt { 1 + 8 ^ { 9 x } } \right)$ .
A) $\sqrt { 9 \cdot \ln 8 \cdot 8 ^ { 9 x } }$
B) $\sqrt { 9 x \cdot 8 ^ { 9 x - 1 } }$
C) $\frac { 9 \cdot \ln 8 \cdot 8 ^ { 9 x } } { 2 \sqrt { 1 + 8 ^ { 9 x } } }$
D) $\frac { 9 x \cdot 8 ^ { 9 x - 1 } } { 2 \sqrt { 1 + 8 ^ { 9 x } } }$

Question

Quizplus · Accepted Answer

To differentiate $\sqrt { 1 + 8 ^ { 9 x } }$, we use the chain rule. Let $u = 1 + 8^{9x}$, then the derivative of the outer function $\sqrt{u}$ with respect to $u$ is $\frac{1}{2\sqrt{u}}$. The derivative of the inner function $u = 1 + 8^{9x}$ with respect to $x$ is $9 \cdot \ln(8) \cdot 8^{9x}$. Applying the chain rule, we multiply these derivatives to get $\frac { 9 \cdot \ln 8 \cdot 8 ^ { 9 x } } { 2 \sqrt { 1 + 8 ^ { 9 x } } }$.

Find ddx(1+89x)\frac { d } { d x } \left( \sqrt { 1 + 8 ^ { 9 x } } \right)dxd​(1+89x​)

Find $\frac { d } { d x } \left( \sqrt { 1 + 8 ^ { 9 x } } \right)$