Hi, I need help resolving a derivative using Definition of Derivative method Hi, I need help resolving a derivative using Definition of Derivative method @Calculus1

Question

anonymous · Answer

the exercise is:   $$\sqrt{9x^{2}+16}$$

anonymous · Answer

this is what I've done:  $$(\sqrt{9x ^{2}+18hx+9h ^{2}+16} - \sqrt{9x ^{2}+16} ) \div h$$

anonymous · Answer

have you tried rationalizing by taking the conugate?

asnaseer · Answer

Using hint from Outkast3r09:$$f(x)=\sqrt {9x ^{2}+16}$$Definition of derivative is$$f\prime(x)=\lim_{h \rightarrow 0}{(f(x+h)-f(x))/h}=\lim_{h \rightarrow 0}{(\sqrt {9(x+h) ^2+16}-\sqrt {9x ^{2}+16})/h}$$$$f\prime(x)=\lim_{h \rightarrow 0}{(\sqrt {9(x+h) ^2+16}-\sqrt {9x ^{2}+16})*(\sqrt {9(x+h) ^2+16}+\sqrt {9x ^{2}+16})/h(\sqrt {9(x+h) ^2+16}+\sqrt {9x ^{2}+16})}$$$$f\prime(x)=\lim_{h \rightarrow 0}{(9(x+h) ^{2}+16-9x ^{2}-16)/h(\sqrt {9(x+h) ^2+16}+\sqrt {9x ^{2}+16})}$$$$f\prime(x)=\lim_{h \rightarrow 0}{(18hx+9h ^{2})/h(\sqrt {9(x+h) ^2+16}+\sqrt {9x ^{2}+16})}$$$$f\prime(x)=\lim_{h \rightarrow 0}{(18x+9h)/(\sqrt {9(x+h) ^2+16}+\sqrt {9x ^{2}+16})}$$$$f\prime(x)=(18x)/2\sqrt {9x ^{2}+16}$$$$f\prime(x)=9x/\sqrt {9x ^{2}+16}$$