Question

I need some help with this calculus 1 question

Answer 1

Step-by-step explanation:

$f(x) = x \sqrt{9 - {x}^{2} } \\ \therefore \: f(x) = \sqrt{9 {x}^{2} - {x}^{4} } \\ \therefore \: f(x) = ({9 {x}^{2} - {x}^{4} } )^{ \frac{1}{2} } \\ differentiating \: both \: sides \: w.r.t. \: x \\ {f}^{'}(x) = \frac{1}{2} \times ({9 {x}^{2} - {x}^{4} } )^{ - \frac{1}{2} } \\ \hspace{30 pt} \times (9 \times 2x - 4 {x}^{3} ) \\ \therefore \: {f}^{'}(x) = \frac{1}{2} \times ({9 {x}^{2} - {x}^{4} } )^{ - \frac{1}{2} } \\\hspace{32 pt}\times (18x - 4 {x}^{3} ) \\ \therefore \: {f}^{'}(x) = \frac{1}{2} \times ({9 {x}^{2} - {x}^{4} } )^{ - \frac{1}{2} } \\\hspace{32 pt} \times 2(9x - 2 {x}^{3} ) \\ \therefore \: {f}^{'}(x) = ({9 {x}^{2} - {x}^{4} } )^{ - \frac{1}{2} } \times (9x - 2 {x}^{3} ) \\ \therefore \: {f}^{'}(x) = \frac{1}{x({9 - {x}^{2} } )^{ \frac{1}{2} } } \times x(9 - 2 {x}^{2} ) \\ \\ \: \: \: \: \: \: \red{ \boxed{ \bold{\therefore \: {f}^{'}(x) = \frac{(9 - 2 {x}^{2} ) }{ \sqrt{{9 - {x}^{2} }}}}}}$