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Asked by Raj Last Modified
As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for finding online coaching tuition. Now, let's dive into differentiating the given function with respect to xx:
f(x)=(ax2+cotx)(p+qcosx)f(x)=(ax2+cotx)(p+qcosx)
To differentiate this function with respect to xx, we'll use the product rule. According to the product rule, if u(x)u(x) and v(x)v(x) are differentiable functions of xx, then the derivative of u(x)⋅v(x)u(x)⋅v(x) with respect to xx is given by:
ddx[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)dxd[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)
Where u′(x)u′(x) and v′(x)v′(x) are the derivatives of u(x)u(x) and v(x)v(x) with respect to xx, respectively.
In our case, u(x)=ax2+cotxu(x)=ax2+cotx and v(x)=p+qcosxv(x)=p+qcosx. Let's differentiate each of these:
u′(x)u′(x): u′(x)=ddx(ax2+cotx)u′(x)=dxd(ax2+cotx) =2ax−csc2(x)=2ax−csc2(x)
v′(x)v′(x): v′(x)=ddx(p+qcosx)v′(x)=dxd(p+qcosx) =−qsinx=−qsinx
Now, applying the product rule:
f′(x)=(2ax−csc2(x))(p+qcosx)+(ax2+cotx)(−qsinx)f′(x)=(2ax−csc2(x))(p+qcosx)+(ax2+cotx)(−qsinx)
And that's the derivative of the given function with respect to xx.
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